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laurmaso
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Homework Statement
A projectile is launched from a cliff that is 84m tall with an initial speed of 32m/s at an angle of 25 degrees above the horizontal. The object is in the air for a total of 5.74s. What is its impact velocity and at what angle?
Homework Equations
Vfy = Voy + ayt
Vf^2 = Vfx^2 + Vfy ^2
tan(theta) = (Vfy)/(Vfx)
The Attempt at a Solution
I have that the answer is 51.7m/s at an angle of 55.9 degrees below the horizontal. However, when I attempt to solve, my answers don't match up:
Vfy = Voy + ayt
= 0 + (-9.8m/s^2)(5.74s)
= -56.25
Vf^2 = Vfx^2 + Vfy^2
= (32)^2 + (-56.25)^2
= 1024 + 3164.06
= 4188.29
Vf = 64.71m/s <------ incorrect
Then when I plug into find the angle of impact:
tan(theta) = (Vfy)/(Vfx)
= (-56.25)/(32)
= -1.7578
tan^-1 (-1.5758) = -60.36 degrees below the horizontal <------- also incorrect.
I'm not sure what I'm doing wrong. Any suggestions??
I also need to calculate the peak height of the projectile; I first used:
t = (Voy)/g
= 32sin25/(-9.8m/s^2)
= 13.5/(-9.8)
= 1.37s
then y = (1/2)at^2 + Voyt
= (-4.9m/s^2)(1.37s)^2 + 13.5m/s(1.37s)
= -9.2 + 18.5
= 9.3m <-------- this answer cannot be correct, not sure what I'm doing wrong
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