You are thinking about taking gymnastics, so you go to the facility and get an i

Question

You are thinking about taking gymnastics, so you go to the facility and get an idea of what to expect by looking out from the viewing room. The viewing room window is 2.90 m above the trampoline directly below, so it is perfect for viewing the the facility. Occasionally someone jumps past the window and then back down. On one occurrence a gymnast went up past the window and came back down; as he passed the window on the way down, you notice that his speed is 11.8 m/s.

What must have been his initial speed coming off the trampoline?

Explanation / Answer

while going down, he will reach the tramploin at same speed , at which he has left the trampolin

so consider the part of the path from when you measured his veloicty as 11.8 m/s till he travels 2.9 m below till the trampolin

then initial speed for our considered path=11.8 m/s

final speed to be calculated

acceleration=9.8 m/s^2

distance =2.9 m

then using the formula:final veloicty^2-initial veloicty^2=2*acceleration*distance

final veloicty=14 m/s

hence he left the trampolin with 14 m/s

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