Posted by **Julia ** on Sunday, February 3, 2013 at 12:41pm.

A projectile is fired at 45.0° above the horizontal. Its initial speed is equal to 42.5 m/s. Assume that the free-fall acceleration is constant throughout and that the effects of the air can be ignored. What is the maximum height reached by the projectile?

At what time after being fired does the projectile reach this maximum height?

- Physics please help!! -
**Damon**, Sunday, February 3, 2013 at 1:23pm
Vi = 42.5 sin 45 = 30 m/s

v = Vi - 9.81 t

0 = 30 - 9.81 t

t = 3.06 seconds to top

h = 0 + Vi t - (1/2)9.81 t^2

h = 30(3.06) - 4.9 (3.06)^2

= 91.8 - 45.9

= 91.7 m

- Physics -
**chikodi john**, Friday, May 29, 2015 at 11:49am
a ball thrown vertically upward from the top of a tower 60m high a velocity of 30ms what is the maximum height above the ground level.how long does it take to reach the ground

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