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?'
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)
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> Please correct , " height achieved in secs. " as " height achieved in 1.3409 secs. " .
Tulsiram Garg Mon, Sep 27, 2021 3:19 AM