Asked by Joe
A rock is thrown upwards (vertically) from the ground with an initial velocity of 40 feet per second. If acceleration due to gravity is -10 feet per second squared, how high will the rock go?
Answers
Answered by
Reiny
You are using the wrong units, the value of -10 is associated with metres not feet. (actually in most cases it is considered -9.8 )
I will assume you meant metres, not feet
height = -5t^2 + 40t
velocity = -10t + 40
at max height, velocity = 0
-10t+40 = 0
t= 4
height = -5(4^2) + 40(4) = 40 m
If your problem stated feet, then the equation would have been
height = -16t^2 + 40t
etc
I will assume you meant metres, not feet
height = -5t^2 + 40t
velocity = -10t + 40
at max height, velocity = 0
-10t+40 = 0
t= 4
height = -5(4^2) + 40(4) = 40 m
If your problem stated feet, then the equation would have been
height = -16t^2 + 40t
etc
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