Posted by Gabriel on .
A balloon, rising vertically with a velocity of 16 feet per second, releases a sandbag at the instant it is 64 feet above the ground.
a. How many seconds after its release will the bag strike the ground?
b. At what velocity will it hit the ground?
calculus - INTEGRATION -
This question does not require calculus or integration.
It does require that you know the acceleration of gravity is 32 ft/s^2.
a. Solve for time t:
height = 64 + 16 t - (32/2)t^2 = 0
t^2 -t -4 = 0
t = (1/2)[1+ sqrt17] = 2.562 seconds
(ignore then negative root, -1.562 s)
b. V2 at impact = 16 - 32*2.562
= -66.0 ft/s
Energy Check: V2^2/2 - V1^2/2 = g*16 ft