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October 1, 2014

October 1, 2014

Posted by **kate** on Sunday, June 19, 2011 at 5:44pm.

(b) What is the ball's maximum height above the ground?

HELP!

- physics -
**Damon**, Sunday, June 19, 2011 at 5:52pmHow high is the wall if a dropped object hits ground in 1.1 seconds?

h = (1/2) g t^2

= 4.9(1.1)^2 = 5.93 meters high

so

u = 54 cos 34 forever

Vi =54 sin 34

v = 54 sin 34 - 9.8 t

h = 5.93 + Vi t - 4.9 t^2

when h = 0, the ball hits ground

0 = 5.93 + (54 sin 34) t - 4.9 t^2

solve for t with quadratic eqn

then

distance = u t = 54 cos 34 * t

for the last part, v = 0 at the top so

9.8 t = 54 sin 34

use that t in

h = 5.93 + (54 sin 34) t - 4.9 t^2

to get h at the top

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