At time t = 0, a rock climber accidentally allows a piton to fall freely from a high point on the rock wall to the valley below him. Then, after a short delay, his climbing partner, who is 11.8 m higher on the wall, throws a piton downward. The positions y of the pitons versus t during the falling are given in the figure. With what speed is the second piton thrown?
From the graph, it is clear that both the pitons reaches ground after t = 3 s
Suppose v is the speed of throwing of the second piton.
h = height of the first piton above the ground.
So, for the first piton the equation is -
h = (1/2)*g*t^2 = (1/2)*g*3^2 [Since, initial speed of the first piton = 0]-------------------------------------(i)
For the second piton -
h + 11.8 = v*t + (1/2)*g*t^2 = v*3 + (1/2)*g*3^2----------------------------------------(ii)
From equation (i) and (ii) -
(h+11.8) - h = [v*3 + (1/2)*g*3^2] - (1/2)*g*3^2
=> 11.8 = v*3
=> v = 11.8/3 = 3.93 m/s (Answer)
At time t = 0, a rock climber accidentally allows a piton to fall freely from...
At time t = 0, a rock climber accidentally allows a piton to fall freely from a high point on the rock wall to the valley below him. Then, after a short delay, his climbing partner, who is 11.8 m higher on the wall, throws a piton downward. The positions y of the pitons versus t during the falling are given in the figure. With what speed is the second piton thrown? 0 t (s)
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At time t = 0 s, apple 1 is dropped from a bridge onto a roadway beneath the bridge; somewhat later, apple 2 is thrown down from the same height. The figure gives the vertical positions y of the apples versus t during the falling, until both apples have hit the roadway. The scaling is set by ts = 2.0 s. With approximately what speed is apple 2 thrown down? 0.5 1.5 2.5 (s)
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