Question

A golfer gives a ball a maximum initial speed of 58.8 m/s. What is the highest tree the ball could clear on its way to the longest possible hole-in-one? Express your answer using three significant figures.

Answer 1

Horizontal range of ball R = Vi^2 sin(2 theta)/g

Where theta is angle of projection. For maximum R, sin(2 theta) must be equal to one or theta = 45 deg

Hence for longest possible hole-in, angle of projection must be

45 deg.

Maximum height of projectile = Vi^2 sin^2(theta) / 2g

Hence maximum height for longest hole-in = (58.8)^2(m^2/s^2)*(1/2)/2*9.8(m/s^2) = 88.1 m = tallest tree ball can clear.