Posted by **Gabriel** on Wednesday, February 2, 2011 at 10:21pm.

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 -
**drwls**, Thursday, February 3, 2011 at 2:03am
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

as required

## Answer this Question

## Related Questions

- Physics - A hotair balloonist, rising vertically with a constant speed of 5.00 ...
- Physics - A hotair balloonist, rising vertically with a constant speed of 5.00 ...
- Physics 1 - A hot-air balloonist, rising vertically with a constant velocity of ...
- Physics - A hot-air balloonist, rising vertically with a constant velocity of ...
- physics - a hot air balloonist, rising vertically with a constant velocity of ...
- Calculus - A balloon is rising vertically above a level, straight road at a ...
- Physics - 1 A hot-air balloon is rising upwards at a constant velocity of 5m.s ...
- Physics - A balloonist, riding in the basket of a hot air balloon that is rising...
- physics - A balloonist, riding in the basket of a hot air balloon that is rising...
- physical science - A hot air balloon is rising upwards at a constant velocity of...