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A box is launched upward from the ground at a speed of 57 feet per second. Calculate the height in feet of the box at 3 seconds. use g=32 ft/s^2 as the acceleration due to gravity

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The vertical height traveled is given by
h = ut +- (1/2)g*t^2
where
u = 57 ft/s, initial speed
t = tme, s
g = 32 ft/s^2

After 3 seconds,
h = (57 ft/s)*(3 s) - (1/2)*(32 ft/s^2)*(3 s)^2
= 171 - 144
= 27 ft

Answer: 27 ft
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