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mg_perez75 · Sep 26, 2011, 09:11 PM

I would like to know the steps to take to solve please.
A golfer can hit a golf ball a horizontal distance of over 300m on a good drive. What maximum height will a 301.5m drive reach if it is launched at an angle of 25degrees to the ground?

Unknown008 · Sep 27, 2011, 02:12 AM

Let v be the velocity the ball leaves the ground.

You can get the vertical and horizontal components of the velocity.

On the horizontal for displacement, you have:
$s_h = u_ht$

On the vertical for displacement, you have:
$s_v = u_vt - \frac12 gt^2$

1. Get t in terms of the horizontal distance and horizontal component of the velocity from the first equation.

2. Substitute this time divided by 2 into the second equation. (If it took 50 s to travel 300 m, the ball would be at its highest position at 25 s)

$s_h$ = 301.5 m
g = 9.81 m/s^2

You can get $s_v$ the maximum vertical displacement.

shqiperona · Nov 16, 2011, 03:03 PM

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