Question

Firemen are shooting a stream of water at a burning building using a high-pressure hose that shoots out the water with a speed of 25.0 m/s as it leaves the end of the hose. Once it leaves the hose, the water moves in projectile motion. The firemen adjust the angle of elevation α of the hose until the water takes 3.00 s to reach a building 45.0 m away. You can ignore air resistance; assume that the end of the hose is at ground level.
a) Find the angle of elevation α
b) Find the speed and acceleration of the water at the highestpoint in its trajectory
c) How high above the ground does the water strike thebuilding, and how fast is it moving just before it hits thebuilding?

Answer 1

Answer:

Answered

Explanation:

The velocity is given by

 V_o cos α × t = 45 m

cos α = 45 / (25)×(3)

cos α=0.6

α = 53.13°

The velocity along x-axis

V_x = 25 cos 53.13 = 15 m/s

 V_y = 0

 V = 15 m/s

acceleration  = 9.8 m/s^2  downward.

c)The height is

y = V_o sin α t - (1/2)g t^2

putting values we get

= 15.9 m

now,

V_y = V_o sin α - g t = - 9.41 m/s

therefore,  resultant velocity V

V= √(Vx^2 +Vy^2 ) =17.7 m/s