Question

the maximum height attained by the projectile motion is 600m and the range is 500m find
a. angle of projection
b. velocity of projection
c. total time of flight

Answer 1

Vertical problem:
at max height (half the trip)
600 = Vi t - .5*9.8 t^2
.5*9.8 t^2 - Vi t + 600 = 0
also
v = 0
so
0 = Vi -gt
t = Vi/g = Vi/9.8
so
.5*9.8(Vi^2/9.8^2) - Vi^2/9.8 +600 = 0
600 = .5*VI^2/9.8
Vi = 108 m/s initial up speed
t = rising time = 108/9.8 = 11.1 s
total time = 22.2 s
u = 500/22.2 = 22.5 m/s

tan theta = Vi/u = 108/22.5
= 4.8
theta = 78.2 degrees

speed = sqrt(108^2+22.5^2)

check my arithmetic !