# Physics question I need some help with...

• Oct 12th 2009, 08:03 PM
M23
Physics question I need some help with...
A water balloon is dropped on some unsuspecting sunbathers from a stationary hot air balloon. The water balloon is 353 m above the beach. If the water balloon accelerates downwards at 9.81 m/sē, how long will it take to hit the beach?

Any help would be much appreciated.
• Oct 12th 2009, 08:42 PM
Gusbob
Use the formula
$s = \frac{1}{2}at^2$

Where s is displacement (height)
a is acceleration
t is time
• Oct 13th 2009, 03:10 AM
BabyMilo
Quote:

Originally Posted by M23
A water balloon is dropped on some unsuspecting sunbathers from a stationary hot air balloon. The water balloon is 353 m above the beach. If the water balloon accelerates downwards at 9.81 m/sē, how long will it take to hit the beach?

Any help would be much appreciated.

Here is an example:

A ball is thrown vertically upwards with speed 20m/s from the top of a tower of height 25m. Find the time until the ball hits the ground

so:
a=-9.8
u=20
t?
v=X
x or s = -25

s=ut+0.5at^2

-25=20t+1/2*(-9.8)*t^2
-25=20t-4.9t^2
4.9t^2-20t-25=0
t=5.08498

The ball hits the ground after 5.08s (3s.f.)

Use this principle, to work out your question!

The answer should work out to be 8.48s (3s.f.) if I've done it correctly.

Please anyone else correct me if im wrong!
• Oct 13th 2009, 12:49 PM
M23
Thanks a lot!