Coin A is thrown the top of a 328ft tower with a speed of 49 ft/s. Coin B is drop from the top of the tower 2 seconds later. How far below the top of the tower does the coin A pass coin B
Coin A is thrown the top of a 328ft tower with a speed of 49 ft/s. Coin B is drop from the top of the tower 2 seconds later. How far below the top of the tower does the coin A pass coin B.
Let the tower’s top coincide with the origin of the coordinate system. All the upward quantities will be positive and the downward, negative.
Let v0A denote coin A’s initial upward speed since had coin A been thrown downwards, it would never cross coin B as coin B is released later.
As coin B is released from rest, its initial speed (v0B) is zero.
Let g denote the gravitational acceleration.
Coin B accelerates downward due to g for 2 seconds less than coin A.
Ignoring air resistance, coin A and coin B will be at the same position (y) at the time (t). This position can be given by using the second kinematical equation as follows:
Equate the above two equations and solve for t as follows:
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