Asked by josephz
A rock is thrown vertically from the edge of a cliff that is 500 ft high. It takes 6.5 seconds for the rock to hit the ground at the base of the cliff. Assume air resistance is negligible.
(a) What was the initial velocity (magnitude and direction) of the rock?
(b) What was the magnitude of the velocity of the rock just before impact with the ground?
(a) What was the initial velocity (magnitude and direction) of the rock?
(b) What was the magnitude of the velocity of the rock just before impact with the ground?
Answers
Answered by
drwls
(a) They have already told you that the initial rock direction was vertical (up).
y = 500 + Vo*t - 16.1 t^2
= 0 then t = 6.5
0 = 500 + 6.5*Vo - 16.1*(6.5)^2
Solve for Vo.
V(t) = Vo - 16.1 t
Substitute t = 6.5 s and solve for V at that time.
The number 16.1 is the value of g/2, in units of ft/s^2
y = 500 + Vo*t - 16.1 t^2
= 0 then t = 6.5
0 = 500 + 6.5*Vo - 16.1*(6.5)^2
Solve for Vo.
V(t) = Vo - 16.1 t
Substitute t = 6.5 s and solve for V at that time.
The number 16.1 is the value of g/2, in units of ft/s^2
Answered by
john
youve got the wrong answer
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