Question

A golf ball is hit horizontally at 40 m/s from the top of a hill that is 2.5 m high. If the terrain around the hill is nearly flat,approximately how far will the golf ball fly? Use - 9.81 m/s2 for the acceleration caused by gravity. Ignore air resistance.Round any intermediate calculations to no less than six decimal places, and round your final answer to two decimal places.

Answer 1

28.56 m

Step-by-step explanation

Let's make a diagram of this situation to understand it better,

The ball has an initial velocity of 40m/s, which is horizontal. This means that there is no vertical initial velocity.

We have to find the horizontal distance the golf ball flies. Since the horizontal velocity is constant - this is because there is no acceleration in that direction, the distance the ball travels is,

`\Delta x=v_(ox)\cdot t`

The horizontal initial velocity is given, but we have to find the time the ball was in the air. To find it, we use the vertical distance the ball travels - which we know is the height of the hill. In this case, we do have vertical acceleration - the acceleration of gravity, so the vertical distance the ball travels, as shown in the diagram, has a parabolic form and it is given by the equation,

`\Delta y=v_(oy)\cdot t+(1)/(2)gt^2`

The initial vertical velocity is zero because the ball is hit horizontally,

`\Delta y=(1)/(2)gt^2`

Solve for t. Multiply both sides by 2/g,

`\begin{gathered} \Delta y(2)/(g)=(1)/(2)g\cdot(2)/(g)\cdot t^2 \\ (2\Delta y)/(g)=t^2 \end{gathered}`

And take the square root to both sides of the equation,

`t=\sqrt[]{(2\Delta y)/(g)}`

Δy is the height of the hill, 2.5m, and g = 9.81m/s²,

`t=\sqrt[]{(2\cdot2.5m)/(9.81m/s^2)}=\sqrt[]{(5m)/(9.81m/s^2)}=\sqrt[]{0.509684s^2}=0.713922s`

This is the time the ball was in the air for. Now we can find the distance it traveled,

`\Delta x=40m/s\cdot0.713922s=28.56m`

The golf ball flew 28.56 m horizontally.