An aircraft having velocity 180 mih leaves its projectile wi

An aircraft having velocity 180 mi/h leaves its projectile with 30° downward with respect to horizontal line (x-axis). The horizontal distance between the point the projectile left and the point where it hits the ground is 701 m. The height at which the aircraft leaves the projectile is

Solution

V = speed of projectile = 180 mi/h = 80.5 m/s

V_ox = initial velocity in X-direction = V cos30 = 80.5 Cos30 = 69.72 m/s

X = distance travelled in X-direction = 701 m

t = time taken

t = X/V_ox = 701 / 69.72 = 10.1 sec

Consider the motion in Y-direction

Y = vertical displacement = height

V_oy = final velocity in Y-direction = - V sin30 = - 80.5 Sin30 = - 40.25 m/s

a = acceleration = - 9.8 m/s²

using the equation

Y = V_oy t + (0.5) a t²

Y = - 40.25 (10.1) + (0.5) (-9.81) (10.1)² = - 906.88 m

height = 906.88 m