Question

a A golfer hits a golf ball at an angle of 25 degrees to the ground. If the golf ball covers a horizontal distance of 301. 5 m, what is the ball's maximum height? (Hint: at the top of its flight, the ball's vertical velocity component will be zero.)

Answer 1

As given by the question

There are given that the golf ball covers a horizontal distance of 301.5 m.

Now,

`\begin{gathered} R=301.5\text{ m} \\ \theta=25^(\circ) \end{gathered}`

The vertical motion is:

`v=u+at`

Where

`\begin{gathered} v=u\sin \theta \\ u=0\text{ m/s} \\ a=-gm/s^(\square) \end{gathered}`

Put the value

`t=(2u\sin \theta)/(g)`

Then,

The range of projectile motion :

`\begin{gathered} R=(u^2\sin ^2\theta)/(g) \\ u^2=393.59g \end{gathered}`

And,

The height

`\begin{gathered} H=(u^2*\sin^2\theta)/(2g) \\ H=\frac{393.59^{}*\sin ^2(25)}{2g} \\ H=35.15\text{ m} \end{gathered}`

Hence, the maximum height that can reach by the ball is 35.15 m