- #1
sirfederation
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1. A rock is thrown from the roof of a building, with an initial velocity of 10 m/s at an angle of 30 degrees above the horizonal. The rock is observed to strike the ground 43 m from the base of the building. What is the height of the building?
This is what I did:
First I set my origin at the top of the building so the height would be in the negative direction
Vxi = 10 cos 30 = 8.66 m/s
Vyi = 10 sin 30 =5 m/s
Xxf - Xxi = 1/2( Vi-Vf)T
43-0 = 1/2 (8.66 + 8.66)T
43 = 8.66T
4.97 s = T
Vyf = 5 m/s -(9.8 m/s^2)(4.97s)
Vyf = -43.61 m/s
Yf-Yi = 1/2(Vi + Vf)T
Yf = 1/2(5 -43.61)4.96
Yf = -95.75 m
What I was wondering is if I found Time which is T and the height correctly?
This is what I did:
First I set my origin at the top of the building so the height would be in the negative direction
Vxi = 10 cos 30 = 8.66 m/s
Vyi = 10 sin 30 =5 m/s
Xxf - Xxi = 1/2( Vi-Vf)T
43-0 = 1/2 (8.66 + 8.66)T
43 = 8.66T
4.97 s = T
Vyf = 5 m/s -(9.8 m/s^2)(4.97s)
Vyf = -43.61 m/s
Yf-Yi = 1/2(Vi + Vf)T
Yf = 1/2(5 -43.61)4.96
Yf = -95.75 m
What I was wondering is if I found Time which is T and the height correctly?
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