Posted by **john** on Thursday, May 2, 2013 at 7:27pm.

A projectile was fired out of a lead ejector at an angle of 45 degrees up from horizontal. The projectile velocity was 1000.00 meters per second. How high did it go? and how far did the projectile go?

- physics -
**Damon**, Thursday, May 2, 2013 at 8:20pm
initial speed up Vi = 1000 sin 45 = 707 m/s

v = Vi - 9.81 t

when v = 0, we go no higher

t = 707/9.81 = 72 seconds upward

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

h = 707 (72) - 4.9 * 5184

= 50904-25427

25,000 m

u = horizontal speed = 1000cos 45 = 707

u t = 707 * 2*72 = 102,000 m

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