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?

Answer 1

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