Maximum range or optimal angle

15 A projectile is launched from a point on level ground with an initial velocity \(u\) at an angle \(\theta\) above the horizontal.

1. Show that the range of the projectile is given by \(\frac { 2 u ^ { 2 } \sin \theta \cos \theta } { g }\).
2. Determine the set of values of \(\theta\) for which the maximum height of the projectile is greater than the range, where \(\theta\) is an acute angle. Give your answer in degrees.

6. A particle \(P\) is projected from a point \(A\) on horizontal ground with speed \(u\) at an angle of elevation \(\alpha\) and moves freely under gravity. \(P\) hits the ground at the point \(B\).

1. Show that \(A B = \frac { u ^ { 2 } } { g } \sin 2 \alpha\). An archer fires an arrow with an initial speed of \(45 \mathrm {~ms} ^ { - 1 }\) at a target which is level with the point of projection and at a distance of 80 m . Given that the arrow hits the target,
2. find in degrees, correct to 1 decimal place, the two possible angles of projection.
3. Write down, with a reason, which of the two possible angles of projection would give the shortest time of flight.
   (2 marks)
4. Show that the minimum time of flight is 1.8 seconds, correct to 1 decimal place.
   (2 marks)

A particle \(P\) is projected with speed \(u\) at an angle \(\theta\) above the horizontal from a point \(O\) on a horizontal plane and moves freely under gravity. The horizontal and vertical displacements of \(P\) from \(O\) at a subsequent time \(t\) are denoted by \(x\) and \(y\) respectively.

1. Use the equation of the trajectory given in the List of formulae (MF19), together with the condition \(y = 0\), to establish an expression for the range \(R\) in terms of \(u\), \(\theta\) and \(g\). [2]
2. Deduce an expression for the maximum height \(H\), in terms of \(u\), \(\theta\) and \(g\). [2]

Fig. 7 shows the trajectory of an object which is projected from a point O on horizontal ground. Its initial velocity is \(40\text{ms}^{-1}\) at an angle of \(\alpha\) to the horizontal. \includegraphics{figure_1}

1. Show that, according to the standard projectile model in which air resistance is neglected, the flight time, \(T\) s, and the range, \(R\) m, are given by $$T = \frac{80\sin\alpha}{g} \text{ and } R = \frac{3200\sin\alpha\cos\alpha}{g}.$$ [6] A company is designing a new type of ball and wants to model its flight.
2. Initially the company uses the standard projectile model. Use this model to show that when \(\alpha = 30°\) and the initial speed is \(40\text{ms}^{-1}\), \(T\) is approximately \(4.08\) and \(R\) is approximately \(141.4\). Find the values of \(T\) and \(R\) when \(\alpha = 45°\). [3] The company tests the ball using a machine that projects it from ground level across horizontal ground. The speed of projection is set at \(40\text{ms}^{-1}\). When the angle of projection is set at \(30°\), the range is found to be \(125\) m.
3. Comment briefly on the accuracy of the standard projectile model in this situation. [1] The company refines the model by assuming that the ball has a constant deceleration of \(2\text{ms}^{-2}\) in the horizontal direction. In this new model, the resistance to the vertical motion is still neglected and so the flight time is still \(4.08\) s when the angle of projection is \(30°\).
4. Using the new model, with \(\alpha = 30°\), show that the horizontal displacement from the point of projection, \(x\) m at time \(t\) s, is given by $$x = 40t\cos 30° - t^2.$$ Find the range and hence show that this new model is reasonably accurate in this case. [4] The company then sets the angle of projection to \(45°\) while retaining a projection speed of \(40\text{ms}^{-1}\). With this setting the range of the ball is found to be \(135\) m.
5. Investigate whether the new model is also accurate for this angle of projection. [3]
6. Make one suggestion as to how the model could be further refined. [1]

A ball is projected forward from a fixed point, \(P\), on a horizontal surface with an initial speed \(u\text{ ms}^{-1}\), at an acute angle \(\theta\) above the horizontal. The ball needs to first land at a point at least \(d\) metres away from \(P\). You may assume the ball may be modelled as a particle and that air resistance may be ignored. Show that $$\sin 2\theta \geq \frac{dg}{u^2}$$ [6 marks]

A ball \(B\) is projected with speed \(V\) at an angle \(\alpha\) above the horizontal from a point \(O\) on horizontal ground. The greatest height of \(B\) above \(O\) is \(H\) and the horizontal range of \(B\) is \(R\). The ball is modelled as a particle moving freely under gravity.

1. Show that
   1. \(H = \frac{V^2}{2g}\sin^2 \alpha\), [2]
   2. \(R = \frac{V^2}{g}\sin 2\alpha\). [3]
2. Hence show that \(16H^2 - 8R_0 H + R^2 = 0\), where \(R_0\) is the maximum range for the given speed of projection. [5]
3. Given that \(R_0 = 200\text{m}\) and \(R = 192\text{m}\), find
   1. the two possible values of the greatest height of \(B\), [2]
   2. the corresponding values of the angle of projection. [3]
4. State one limitation of the model that could affect your answers to part (iii). [1]

It is given that the trajectory of a projectile which is launched with speed \(V\) at an angle \(\alpha\) above the horizontal has equation $$y = x\tan\alpha - \frac{gx^2}{2V^2}(1 + \tan^2\alpha),$$ where the point of projection is the origin, and the \(x\)- and \(y\)-axes are horizontal and vertically upwards respectively.

1. Express the above equation as a quadratic equation in \(\tan\alpha\) and deduce that the boundary of all accessible points for this projectile has equation $$y = \frac{1}{2gV^2}(V^4 - g^2x^2).$$ [4]
2. A stone is thrown with speed \(\sqrt{gh}\) from the top of a vertical tower, of height \(h\), which stands on horizontal ground. Find
   1. the maximum distance, from the foot of the tower, at which the stone can land, [3]
   2. the angle of elevation at which the stone must be thrown to achieve this maximum distance. [3]