Projectile motion of water hose

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A fireman directs a water stream at a 34.0° angle from a distance of 52.0 m, with an initial speed of 40.0 m/s. The vertical component of the velocity, calculated as 22.37 m/s, is used to find the time of flight. The time is incorrectly calculated using the vertical motion equation; instead, it should be determined using horizontal motion. After correcting the time calculation, the height at which the water strikes the building is found to be approximately 22.7 m. The solution highlights the importance of correctly applying kinematic equations in projectile motion problems.
mandy9008
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Homework Statement


A fireman 52.0 m away from a burning building directs a stream of water from a ground-level fire hose at an angle of 34.0° above the horizontal. If the speed of the stream as it leaves the hose is 40.0 m/s, at what height will the stream of water strike the building?


Homework Equations


Voy=Vo sin θ
t= Voy / g
y= Voy t - 1/2 g t2


The Attempt at a Solution


Voy= 40.0 m/s sin 34.0
Voy= 22.37 m/s

t= 22.37 m/s / 9.8 m/s2
t= 2.28s

y= 22.37 m/s (2.28s) - 1/2 (9.8 m/s2)(2.28s)2
y= 25.53 m
 
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Your solution method is pretty much correct except for your calculation of the time. The equation you've used is t = v_{0y}/g, which is not valid. You can use t = x/v_x, however since the horizontal motion has no acceleration. That should do the trick.
 
22.7 m thanks!
 
The book claims the answer is that all the magnitudes are the same because "the gravitational force on the penguin is the same". I'm having trouble understanding this. I thought the buoyant force was equal to the weight of the fluid displaced. Weight depends on mass which depends on density. Therefore, due to the differing densities the buoyant force will be different in each case? Is this incorrect?

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