Question

A projectile hits a target placed on a level field. It has arange of 37.0 m. The nagle of launch is 33.4 degrees above the horizontal. What was the initial speed?

Answer 1

The horizontal component of velocity was constant so
(V cos 33.4)t = 37
V t = 44.3
V = 44.3/t

in the up direction
y = 0 + V sin 33.4 t - 4.9 t^2
0 = 0 + 24.38 =4.9 t^2
4.9 t^2 = 24.38
t^2 = 4.98
t = 2.23seconds in air
V = 44.3/2.23
V = 19.9 m/s