Question

A football punter wants to kick the ball so that it is in the air for 4.2 s and lands 55 m from where it was kicked. Assume that the ball leaves 1.0 m above the ground.
Part A: At what angle should the ball be kicked?
Part B: With what initial speed should the ball be kicked?

Answer 1

Answer:

$57.24^{\circ}$

24.21 m/s

Explanation:

x = Displacement in x direction = 55 m

y = Displacement in x direction = 0

$y_0$ = Height of the ball when the ball is kicked = 1 m

t = Time taken = 4.2 s

$a_y=g$ = Acceleration due to gravity = $-9.81\ \text{m/s}^2$

u = Initial velocity of ball

Displacement in x direction is given by

$x=u_xt+\dfrac{1}{2}a_xt^2\\\Rightarrow 55=4.2u_x+0\\\Rightarrow u_x=\dfrac{55}{4.2}\\\Rightarrow u\cos\theta=13.1\ \text{m/s}\\\Rightarrow u=\dfrac{13.1}{\cos\theta}$

Displacement in y direction is given by

$y=y_0+u_yt+\dfrac{1}{2}a_yt^2\\\Rightarrow 0=1+4.2u\sin\theta+\dfrac{1}{2}\times (-9.81)\times 4.2^2\\\Rightarrow 0=4.2u\sin\theta-85.5242\\\Rightarrow u\sin\theta=20.36\ \text{m/s}\\\Rightarrow u=\dfrac{20.36}{\sin\theta}$

$\dfrac{13.1}{\cos\theta}=\dfrac{20.36}{\sin\theta}\\\Rightarrow \dfrac{20.36}{13.1}=\dfrac{\sin\theta}{\cos\theta}\\\Rightarrow \theta=\tan^{-1}\dfrac{20.36}{13.1}\\\Rightarrow \theta=57.24^{\circ}$

The ball should be kicked at an angle of $57.24^{\circ}$

$u=\dfrac{13.1}{\cos\theta}=\dfrac{13.1}{\cos57.24^{\circ}}\\\Rightarrow u=24.21\ \text{m/s}$

The initial speed of the ball is 24.21 m/s.