Question

A firefighter directs a stream of water toward a burning building from the ground level. The stream of water leaves the ground with Velocity Vi, given that Vi=21.0 m/s and theta initial= 37.5 degrees and the building is 22.5 m away from the release point: How far above the ground is the stream of water hits the building?

Answer 1

Let us assume the upwards direction as positive and the downwards direction as negative.

Gravitational acceleration = g = -9.81 m/s²

Initial velocity of the stream of water = V₁ = 21 m/s

Angle of the stream of water = = 37.5^o

Initial horizontal velocity of the stream of water = V_1x = V₁Cos = (21)Cos(37.5) = 16.66 m/s

Initial vertical velocity of the stream of water = V_1y = V₁Sin = (21)Sin(37.5) = 12.78 m/s

Distance of the building from the release point = R = 22.5 m

Time taken for the water to reach the building = T

There is no horizontal force on the water therefore the horizontal velocity of the water remains constant.

R = V_1xT

22.5 = (16.66)T

T = 1.35 sec

Height at which the water hits the building = H

H = V_1yT + gT²/2

H = (12.78)(1.35) + (-9.81)(1.35)²/2

H = 8.31 m

Height at which the water hits the building = 8.31 m