A golfer is enjoying a day out on the links.

Question

What maximum height will a 334.5 m drive reach if it is launched at an angle of 18.0 ◦ to the ground? The acceleration due to gravity is 9.81 m/s 2 . Answer in units of m

Answer 1

If we define the time to reach the top as t, and the launch velocity at v, we can find the range which is 334.5

334.5=vcos(18)(2t)

the vertical velocity at launch is vsin(18) and the time to reach the top is

t= vsin(18)/g

Substituting we get

335.4=2(v^2)sin(18)cos(18)/g

solving fo v the only unknown we get

v= 74.8 m/s

Knowing the angle is 18 deg, the vertical component is 74.8 sin(18) = 23.1 m/s

At the top, the very velocity = 0 we can use

0=23.1^2-2(9.81)(h) to find ht

h= 27.2 m