A student standing on the ground throws a ball straight up. The ball leaves the student's hand with a speed of 15m/s when the hand is 2.0m above the ground. How long is the ball in the air before it hits the ground? (The student moves her hand out of the way.)

Question

anonymous · Answer

I wrote a long explanation, which then got lost!

time to top of trajectory, t1:

v = 0 = (15 m/s) - (9.8 m/s^2) t1

height above point of release

h' = (15 m/s) t1 - (9.8 m/s^2)(t1)^2

actual height h = 2 m + h'

time to fall from h is t2 which you get from

-h = -(1/2)(9.8 m/s^2) (t2)^2

total time = t1 + t2