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Answer :
−29.27m/s is the speed of the rock just before hitting the ground when a rock is thrown downward from the top of a 37.6-m-tall tower with an initial speed of 11 m/s. assuming negligible air resistance.
What do you mean by initial speed?
When gravity first exerts force on an object, its initial velocity describes how quickly the object moves. The final velocity, on the other hand, is a vector number which gauges a moving body's speed and direction after it has reached its maximum acceleration.
To find: the speed of the rock just before hitting the ground
v f2 =v i2 +2aΔy, with v i =−11m/s and Δy=−37.6m
vf2 =v i2+2aΔy
v2=(−11m/s) 2+2(−9.80m/s 2)(−37.6m)
v=−29.27m/s
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