Question

A golfer tees off and hits a golf ball at a speed of 31 m/s and at an angle of 35 degrees. What is the maximum height attained by the ball? Round the answer to the nearest meter.

Answer 1

Let's write the law of motions of the ball on both x and y axis. The motion of the ball is a uniform motion on the x-axis, with constant velocity, and uniformly accelerated motion on the y-axis, with constant acceleration g:

`x(t)= v_0 \cos \alpha t`

`y(t)=v_0 \sin \alpha t - (1)/(2) gt^2`
where
`v_0 = 31 m/s` is the initial speed of the ball and
`\alpha=35 ^(\circ)` is the angle at which the ball is launched.

The ball reaches its maximum height when its vertical velocity is zero. The law for the vertical velocity is:

`v_y(t)= v_0 \sin \alpha - gt`
By requiring
`v_y(t)=0`, we find the time t at which the ball reaches the maximum height:

`0=v_0 \sin \alpha -gt`

`t= (v_0 \sin \alpha)/(g) = ((31 m/s)(\sin 35^(\circ)))/(9.81 m/s^2)=1.81 s`

And if we substitute this time t inside the law of motion on the y-axis, y(t), we find the maximum height of the ball:

`h=v_0 \sin \alpha t - (1)/(2) gt^2=(31 m/s)(\sin 35^(\circ))(1.81 s)- (1)/(2)(9.81 m/s^2)(1.81s)^2=`

`=16.11 m`