A ball is thrown vertically upward from a window that is 3.6mabove the ground. I

Question

A ball is thrown vertically upward from a window that is 3.6mabove the ground. Its initial speed is 2.8m/s. How long after the first ball is thrown should a second ballbe simply dropped from the same window so that both balls hit theground at the same time? Please show your steps, thanks! :) A ball is thrown vertically upward from a window that is 3.6mabove the ground. Its initial speed is 2.8m/s. How long after the first ball is thrown should a second ballbe simply dropped from the same window so that both balls hit theground at the same time? Please show your steps, thanks! :)

Explanation / Answer

the time of fall of the second ball is t = (2h/g )                                                         = ( 2* 3.6m / 9.8 m/s2 )                                                         =0.857 s the time of flyght of the first ball is t ' = 2 u/g                                                            =0.53 s as the time of flyght of the first ball is less thanthe time of fall fo the second ball the second ball isdropped before the throwing of the first ball

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