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

$t=\frac{2u\sin \theta}{g}$

Then,

The range of projectile motion :

$\begin{gathered} R=\frac{u^2\sin ^2\theta}{g} \\ u^2=393.59g \end{gathered}$

And,

The height

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

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