Posted by **AMY** on Thursday, July 1, 2010 at 12:09am.

A certain freely failing object requires 1.70 s to travel the last 20.5 m before it hits the ground. From what height above the ground did it fall?

- Physics -
**drwls**, Thursday, July 1, 2010 at 3:15am
Let H be the initial height.

Let T be the time it takes to fall the full height H.

g/2 T^2 = 4.9 T^2 = H

(g/2)(T-1.7)^2 = 4.9(T-1.7)^2 = H - 20.5

Combine the equations to eliminate H, so you can solve for T. Then calculate H.

4.9T^2 = 4.9(T-1.7)^2 + 20.5

4.9T^2 = 4.9T^2 -16.66 T +14.161 +20.5

16.66T = 34.66

T = 2.08 s

H = 4.9 T^2 = 21.2 m

- Physics -
**Maddy**, Saturday, October 23, 2010 at 3:15pm
Find the height from which you would have to drop a ball so that it would have a speed of 6.4 m/s just before it hits the ground.

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