A golfer tees off and hits a golf ball at a speed of 31 m/s and at an angle of 35 degrees. How long was the ball in the air? Round the answer to the nearest tenth of a second.

Answer 1

In projectile motion,
h,max = vo^2 sin^2 (x) / 2g
where
h,max = maximum height
vo = initial velocity
x = angle of release
g = acceleration due to gravity = 9.8 m/s^2

Solving for the maximum height,
h,max = 31^2 * sin^2 (35) / 2(9.8)
h,max = 316.16 / 19.6
h,max = 16.13 m

We can now solve for t using the equation:
h = vo,y*t - (1/2)gt^2
Substituting,
16.13 = 31*sin(35) t - 4.9 t^2
4.9t^2 - 17.78t + 16.13 = 0
This is obviously a quadratic equation, and so we use quadratic formula. getting the roots, we'll have
t = 1.81 s
This time is the time the object reaches the maximum height. Since the problem asks for the time the object is in air, we multiply is by 2 to account for the time it falls back the ground.
t = 3.62 s

Actually, there's really a formula to solve for time of flight, which is
t = 2 vo sin(x) / g
If you use this, you should get approximately the same answer. What I did above is the longer method.

hope this helps~ `u`

Answer 2

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