Question

A fireman d = 49.0 m away from a burning building directs a stream of water from a ground-level fire hose at an angle of θi = 24.0° above the horizontal as shown in the figure. If the speed of the stream as it leaves the hose is vi = 40.0 m/s, at what height will the stream of water strike the building?'

Answer 1

The horizontal range is x=(v; cos 24°) 0.49 = {(40.0) cos 24") t = 0.0134 s The height will the stream of water strike the building is y= yo +(v; sin 24°) = (0)+((40.0) sin 24°)(0.0134) – (9.80)(0.0134)* = = 0.217 m

Answer 2

SOLUTION :

Hose at ground level shoots water at an angle of 24º and speed of 40 m/s. 

upward.

So, 

horizontal speed  of water, Vh = 40 cos (24) = 36.5418 m/s

vertical speed  of water , Vv = 40 sin (24) = 16.269466 m/s

Building is 49 m away from the fire hose.

Time needed to reach the building

= Distance / Horizontal Speed 

= 49 / 36.5418

= 1.3409 sec.

In the vertical direction gravity will decelerate the vertical speed of water

at the deceleration rate of 9.8 m/sec^2 .

So, 

Height achieved in t secs. = u t - 1/2 g t^2  ; (u = Vv)

So, height achieved in  secs. 

= (16.269466)(1.3409) - 1/2 * 9.8 * (1.3409)^2

= 13.0    meters.

Hence, water will hit the 49 meter away building at a height of 13.0 meter.

(ANSWER)