A projectile is shot horizontally ( x direction) from the edge of a vertical cliff 100 m above level ground. Its initial speed is 75 m/s. Neglecting air resistance. At what angle is the velocity vector measured clockwise from the x axis the instant before it hits the ground

Theta = 30.56degree

Explanation:

the angle is θ = -0.533 rad

Explanation:

from the equation of vertical motion

vy²=vy₀² + 2*g*y

where

g= gravity =9.8 m/s²

y= height of the cliff = 100 m

then since the projectile has initially no component of velocity in the vertical direction , vy₀=0

then

vy=√(0² + 2*9.8 m/s²* 100 m) = 44.27 m/s

then the angle θ of the velocity vector  will be

tan θ = -vy/vx

θ = tan ⁻¹[(-44.27 m/s)/75 m/s ]= -0.533 rad

then the angle is θ = -0.533 rad

the force on the thing and the distance the force moves the thing