A lammergeier, flying horizontally at \({\rm{V m}}{{\rm{s}}^{ - 1}}\) releases a bone that hits a rock on the ground 90 metres away, measured horizontally. The bone hits the rock at an angle of \({\rm{4}}{{\rm{5}}^0}\) to the vertical. Assume that, \(t\) seconds after release, the position of the bone is given by \(x = Vt\) and \({\rm{y}} = - 5{{\rm{t}}^2}\) The speed \(V\) of the lammergeier is ?

A projectile, of initial speed \({\rm{V m}}{{\rm{s}}^{ - 1}}\) , is fired at an angle of elevation \(\theta \)  from the origin O . You may assume that the projectile's trajectory is defined by the equations

\(x = Vt\cos \theta \) and \({\rm{y}} = - \frac{1}{2}{\rm{ g}}{{\rm{t}}^2}{\rm{ }} + Vt\sin \theta \)

where \(x\) and \(y\) are the horizontal and vertical displacements of the projectile in metres at time \(t\) seconds after firing , and \(g\) is the acceleration due to gravity. How long does the projectile take to reach its maximum height?

A projectile, of initial speed \({\rm{V m}}{{\rm{s}}^{ - 1}}\) , is fired at an angle of elevation \(\theta \) from the origin O. You may assume that the projectile's trajectory is defined by the equations

\(x = Vt\cos \theta \) and \({\rm{y}} = - \frac{1}{2}{\rm{ g}}{{\rm{t}}^2}{\rm{ }} + Vt\sin \theta \)

where \(x\) and \(y\) are the horizontal and vertical displacements of the projectile in metres at time \(t\) seconds after firing , and \(g\) is the acceleration due to gravity. The range of the projectile is?

A projectile is fired from the origin \(O\) with limited speed, \(V\), in metres per second and an angle of projection \(\alpha\). The time is measured in seconds. The path of the projectile is given by the equations

\[x = V t\cos\alpha\]

\[y = Vt\sin\alpha - \frac{1}{2}gt^2\]

where \(g\) is the acceleration due to gravity.

i) In terms of \(V\) and \(\alpha\), find the maximum height \(h\) attained by the particle.

ii) Find the range, \(R\), in terms of \(V\) and \(\alpha\).

iii) Show that \[\tan\alpha = \frac{4h}{R}.\]

A particle is projected to just clear two walls of height \(h\) metres at distance \(x_2\) and \(x_1\) metres from the point of projection. Given that \(x = V\cos (\alpha) t\) and \(y = V\sin(\alpha)t - \frac{1}{2}gt^2\) where \(V\) is the initial velocity and \(t\) is the time in seconds,

i) Show that \(y = x\tan \alpha - \dfrac{gx^2}{2V^2}\sec^2 \alpha.\)
ii) Hence, or otherwise show that \(\tan \alpha = \dfrac{h(x_1+x_2)}{x_1x_2}.\)

A stone is thrown with an initial velocity \(V_s\), at an angle of projection \(\theta\) so that it will hit a bird, perched at the top of a pole, of height \(h\) metres. At the instant the stone is thrown, the bird flies away in a horizontal straight line at a constant speed of \(V_b\) \(ms^{-1}\). The stone reaches a height of \(2h\) metres, and in descent, touches the bird. You are given that \(y = Vt\sin\theta - \frac{1}{2}gt^2\), where \(g\) is the acceleration due to gravity in \(ms^{-2}\).

i) Show that the maximum height of the stone is obtained when \(V\sin\theta = 2\sqrt{gh}.\)
ii) Show that when the stone reaches a height \(h\) then \(gt^2 - 4\sqrt{gh}t+2h = 0.\)
iii) Show that the time for the stone to reach the top of the pole is \((2-\sqrt{2})\sqrt{\frac{h}{g}} \text{ seconds.}\)
iv) Show that the distance the bird flies is \(V_b(2+\sqrt{2})\sqrt{\dfrac{h}{g}}\) metres.
v) Show that the ratio of the horizontal component of stones velocity \((V_{sx})\) to the bird's velocity is given by \(V_{sx} : V_b = \sqrt{2}+1 : 2.\)

A body is projected so that on its upward path it passes through a point situated \(x_1\) metres horizontally and \(h\) metres vertically from the point of projection. The range is \(R\) metres and the angle of projection is \(\alpha\). The position of the particle is given by

\[x = Vt\cos\alpha \qquad y = Vt\sin\alpha -\frac{1}{2}gt^2.\]

Where \(t\) is time, in seconds, after firing, and \(g\) is the acceleration due to gravity.

i) Show that the path of the trajectory is \(y = x\tan\alpha - \dfrac{gx^2}{2V^2\cos^2\alpha}.\)
ii) Show that \(V^2\cos^2\alpha = \dfrac{gR}{2\tan\alpha}.\)
iii) Show that \(\tan\alpha = \dfrac{Rh}{x_1(R-x_1)}.\)

A particle is projected from \(O\) at an angle of \(\theta\) with a velocity \(V\) metres per second. The particle lands at a distance of \(R\) metres from the origin at time \(T\). The equations of motion of the particle are \(\ddot{x} = 0\) and \(\ddot{y} = -g\).

i) Using calculus, derive the expressions for the position of the particle and time \(t\).
ii) Show that \(\displaystyle \tan\theta = \frac{gT^2}{2R}\).
iii) Show that \(\displaystyle R = \frac{V^2\sin2\theta}{g}\).

Two projectiles are fired from the origin with the same initial speed, \(V\), in metres per second and the angle of projection \(\alpha\) and \(\beta\). Both arrows land \(R\) metres from the point of projection. The acceleration due to gravity \(g\) is assumed to be 10\(ms^{-2}\). The cartesian path of the projectile with a projection angle \(\alpha\) is given by

\[y =x\tan\alpha -\frac{5x^2}{V^2}\sec^2\alpha\]

i) Write down the cartesian path of the projectile with a projection angle of \(\beta\).
ii) Show that the range \(R\) of the particle with projection angle \(\alpha\) is given by \(R = \frac{V^2}{10}\sin(2\alpha)\)
iii) Hence show that \(\alpha+\beta = \dfrac{\pi}{2}\).

Two archers at the same location shoot simultaneously and hit a target at different times. They shoot at an initial velocity of 20\(\text{ms}^{-2}\) but at different angles of elevation. One archer's arrow had an angle of elevation of \(\dfrac{\pi}{6}\). Assume that \(g = 10\text{ms}^{-2}\).

i) Find the distance of the target from the archers.
ii) Find the angle of elevation of the other archer's arrow.
iii) Find the exact time which elapses between the fall of the two arrows.

A projectile is fired with initial velocity \(V\text{ms}^{-1}\) at an angle of projection \(\theta\) from a point \(O\) on horizontal ground. After 2 seconds it just passes over a 4 metre high wall that is 24 metres from the point of projection. Assume acceleration due to gravity is \(10\text{ms}^{-2}\). Assume the equations of displacement are

\[x = Vt\cos\theta\text{ and } y = Vt\sin\theta - 5t^2\]

i) Find \(V\) and \(\theta\).
ii) Find the time taken for the projectile to attain its maximum height.
iii) Find the range of the projectile.

A vertical pole \(PQ\) of height \(h\) metres is \(a\) metres from the point of projection \(O\), such that \(\tan\theta = \dfrac{h}{a}\). Two particles are projected at the same instant from \(O\) in directions making angles \(\alpha_1\) and \(\alpha_2\) with the horizontal in that the former strikes the top of the pole \(P\) at the instant the latter strikes the bottom \(Q\). Assume acceleration due to gravity is 10\(\text{ms}^{-2}\) and assume the equations of displacement are \(x = Vt\cos\alpha\) and \(y = Vt\sin\alpha - 5t^2\).

i) Given that the velocity of the particle with projection angle \(\alpha_1\) is \(V_1\) and the velocity of the particle with projection angle \(\alpha_2\) is \(V_2\), show that \(V_1\cos\alpha_1 = V_2\cos\alpha_2\).
ii) Prove that \(\tan\theta = \tan\alpha_1 - \tan\alpha_2\)

\(O\) and \(R\) are two points \(d\) metres apart on a horizontal ground. A rocket is projected from \(O\) with speed \(V_1\text{ms}^{-1}\) at an angle \(\theta\) above the horizontal, where \(0 < \theta < \frac{\pi}{2}\). At the same instant, another rocket is projected vertically from \(R\) with speed \(V_2\text{ms}^{-1}\). The two rockets move in the same vertical plane under gravity where the acceleration due to gravity is \(g\,\text{ms}^{-2}\). After time \(t\) seconds, the rocket from \(O\) has horizontal and vertical displacements \(x\) metres and \(y\) metres respectively from \(O\), while the rocket from \(R\) has vertical displacement \(Y\) metres from \(R\). The two rockets collide after \(T\) seconds.

i) Given that from \(O\) the equations of displacement are \(x = V_1 t \cos\theta\) and \(y = V_1 t \sin\theta -\dfrac{1}{2}gt^2\), write down the equation of displacement for the rocket at \(R\).
ii) Show that \(d = V_1 T \cos\theta\) and \(V_2 = V_1\sin\theta\)
iii) Explain why \(V_1 > V_2\) for the rockets to collide.
iv) Show that \(T = \dfrac{d}{\sqrt{{V_1}^{2} - {V_2}^{2}}}\)
v) If the two rockets collide at the highest points of their respective flights, show that \(d = \dfrac{V_2\sqrt{{V_1}^{2} - {V_2}^{2}}}{g}\)

The diagram shows an inclined plane that makes an angle of \(\frac{\pi}{6}\) radians with the horizontal. A projectile is fired from \(O\) at the bottom of the incline, with a speed of \(V\text{ms}^{-1}\) at an angle of elevation of \(\frac{\pi}{3}\) to the horizontal as shown.

i) Show that the position of the projectile is given by \(x = \frac{Vt}{2} \text{ and } y = \frac{V\sqrt{3}}{2}t - \frac{1}{2}gt^2\), where \(t\) is the time, in seconds, after firing and \(g\) is the acceleration due to gravity.
ii) Assuming that \(\dfrac{2V^2}{g} = 1\), show that the path of the trajectory of the particle is \(y = x\sqrt{3} - 4x^2\)
iii) Show that the range of the projectile \(r = OT\) metres up the inclined plane is \(\frac{1}{3}\) metres.
iv) Show that, for this trajectory, the initial direction is perpendicular to the direction at which the projectile hits the inclined plane.

A projectile is fired with initial velocity \(V\,\text{ms}^{-1}\) at an angle of \(\theta\) from a point \(O\) on horizontal ground. After 4 seconds it just passes over a 6 metre high wall that is 30 metres from the point of projection. Assume acceleration due to gravity is \(10\,\text{ms}^{-2}\). Assume the equations of displacement are \(x = Vt\cos\theta\) and \(y = Vt\sin\theta - 5t^2\).

i) Find \(V\) and \(\theta\).
ii) Find the time taken for the projectile to attain its maximum height.
iii) Find the range of the projectile.

A particle is projected from a point \(O\) with speed \(30\,\text{ms}^{-1}\) at an angle of elevation \(\displaystyle \tan^{-1}\left(\frac{3}{4}\right)\). One second later, a second particle is projected from \(O\) and it collides with the first particle two seconds after leaving \(O\). Assume the equations of displacement are \(x = Vt\cos\theta\) and \(y = Vt\sin\theta - 5t^2\).

i) Find the angle of elevation of the second particle.
ii) Find the initial velocity of the second particle.

A particle is projected with speed \(V\) to strike at right angles a plane through the point of projection inclined at \(45^\circ\) to the horizontal.

i) Show that the angle of projection is \(\tan^{-1} 3\).
ii) Show that \(R = \dfrac{2\sqrt{2}V^2}{5g}\)

A lammergeier, flying horizontally at \({\rm{25 m}}{{\rm{s}}^{ - 1}}\) releases a bone that hits a rock on the ground. The lammergeier is flying at a height of 180 metres. How long will the bone take to hit the rock? Take \(g = 10\,\,m{s^{ - 2}}\)

A particle is projected so that at any time \(t\) its position \({\rm{(}}x,y{\rm{)}}\) is given by \(x = 36t\), \(y = 15t - \frac{1}{2}g{t^2}\) when the components are measured in meters and time in seconds \({\rm{(g = 10m}}{{\rm{s}}^{ - 2}})\). Find the greatest height reached.

A missile is fired at  \({\rm{100 m}}{{\rm{s}}^{ - 1}}\) to strike, on the ground, a target 500 m away. At what angles could it be launched from? Take \(g = 10m{s^{ - 1}})\)

A projectile, of initial speed \({\rm{V m}}{{\rm{s}}^{ - 1}}\) is fired at an angle of elevation \( \theta \) from the origin O. The projectile's trajectory is defined by the equations. 

\(x = Vt\cos \theta {\rm{ }}\) and \({\rm{y}} = - \dfrac{1}{2}{\rm{ g}}{{\rm{t}}^2}{\rm{ }} + Vt\sin \theta \) ,

where \(x\) and \(y\) are the horizontal and vertical displacements of the projectile in metres at time \(t\) seconds after firing , and \(g\) is the acceleration due to gravity. The range, \(R\), in terms of \(V\) and \(\theta \) is?

A projectile is fired from the origin \(O\) with limited speed, \(V\), in meters per second and an angle of projection \(\alpha \). The time is measured in seconds. The path of the projectile is given by the equations

\[x = Vt\cos \alpha \]

\[y = Vt\sin \alpha - \frac{1}{2}g{t^2}\]

where \(g\) is the acceleration due to gravity.

(i) In terms of \(V\) and \(\alpha \), find the maximum height \(h\) attained by the particle. (3 marks)

(ii) Find the range, \(R\), in terms of of \(V\) and \(\alpha \). (2 marks)

A particle is projected so that at any time \(t\) the equation of the trajectory is given by:

\(x=15t\), \(y=20t-5t^2\) (components measured in metres), find

(i) The initial velocity of the particle (2 marks)

(ii) The maximum height of the particle (1 mark)

(iii) The range of the particle (1 mark)

A particle is projected from a point \(O\) on level horizontal ground with a speed of \(21 \mathrm{~m} \mathrm{~s}^{-1}\) at an angle \(\theta\) to the horizontal. At time \(T\) seconds, the particle passes through the point \(B(12,2)\)

Neglecting the effects of air resistance, the equations describing the motion of the particle are:

\(\begin{gathered}
x=V t \cos \theta \\
y=V t \sin \theta-\dfrac{1}{2} g t^2
\end{gathered}\)

where \(t\) is the time in seconds after projection, \(g \mathrm{~m} \mathrm{~s}^{-2}\) is the acceleration due to gravity where \(g=9.8 \mathrm{~m} \mathrm{~s}^{-2}\) and \(x\) and \(y\) are measured in metres. Do NOT prove these equations.

(i) By considering the horizontal component of the particle's motion, show that \(T=\dfrac{4}{7} \sec \theta\)

(ii) By considering the vertical component of the particle's motion and, using the result from part (a) (i), show that \(4 \tan ^2 \theta-30 \tan \theta+9=0\).

(iii) Find the particle's least possible flight time from \(O\) to \(B\). Give your answer correct to two decimal places.