Basic trajectory calculations

Andrew hits a tennis ball vertically upwards towards his sister Barbara who is leaning out of a window 7.5 m above the ground to try to catch it. When the ball leaves Andrew's racket, it is 1.9 m above the ground and travelling at \(21 \text{ m s}^{-1}\). Barbara fails to catch the ball on its way up but succeeds as the ball comes back down. Modelling the ball as a particle and assuming that air resistance can be neglected,

1. find the maximum height above the ground which the ball reaches. [4 marks]
2. find how long Barbara has to wait from the moment that the ball first passes her until she catches it. [6 marks]

A ball of mass \(m\) is thrown vertically upwards from the ground with an initial speed \(u\). When the speed of the ball is \(v\), the magnitude of the air resistance is \(mkv\), where \(k\) is a positive constant. By modelling the ball as a particle, find, in terms of \(u\), \(k\) and \(g\), the time taken for the ball to reach its greatest height. [8]

In this question use \(g = 9.8\) m s\(^{-2}\) A toy shoots balls upwards with an initial velocity of 7 m s\(^{-1}\) The advertisement for this toy claims the balls can reach a maximum height of 2.5 metres from the ground.

1. Suppose that the toy shoots the balls vertically upwards.
   1. Verify the claim in the advertisement. [2 marks]
   2. State two modelling assumptions you have made in verifying this claim. [2 marks]
2. In fact the toy shoots the balls anywhere between 0 and 11 degrees from the vertical. The range of maximum heights, \(h\) metres, above the ground which can be reached by the balls may be expressed as $$k \leq h \leq 2.5$$ Find the value of \(k\) [4 marks]

A particle is projected with speed \(U \text{ m s}^{-1}\) at an angle \(\theta\) above the horizontal, where \(\sin \theta = \frac{12}{13}\), and reaches its maximum height after \(2.4\) seconds.

1. Find \(U\) and the maximum height reached by the particle. [4]
2. Find the horizontal range of the particle. [4]