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:

for vertical component (Vy) of the projectile; given s = 100m, u = 0m/s, Vx = 75m/s

from the second equation of motion; Vy^2 = U^2 + 2as V^y2 = 0 + 2 x 9.81 x 100 Vy = square root (1962) = 44.29m/s

Angle between them theta = arctan (Vy/vx) = arctan (44.29/75) Theta = arctan (0.59059) = 30.56 degree

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