A golfer, standing on a fairway, hits a shot to a green that is elevated 5.50 m above the point where she is standing. The ball leaves her club at an angle of 37.0° above the ground, with a speed of 31.1 m/s.

Question

A golfer, standing on a fairway, hits a shot to a green that is elevated 5.50 m above the point where she is standing. The ball leaves her club at an angle of 37.0° above the ground, with a speed of 31.1 m/s. 
Find the time that the ball is in the air before it hits the green. (in s)
How far did the ball travel horizontally? (in m)
Determine the ball's speed right before it hits the green. (in m/s)

Steve · Answer

the height of the ball as a function of time is

y = 31.1*sin37.0°*t - 4.9t^2
= 18.7t - 4.9t^2

So, you want to find t when y = 5.5 (and is on its way down). t=3.5 s

Since the horizontal speed is constant, just multiply it by 3.5 to find the distance.

v = 18.7 - 9.8t
and plug int=3.5 to find the vertical speed when it hits. You need to factor in the horizontal speed, as well, for the total speed.

Anonymous · Answer

OMG, I have looked at the other solutions for this same question and they were all wrong, until I got to yours! Thanks, Steve!