Question

At time t =0, a rock climber accidentally allows a piton (metal spike) 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 10 m higher on the wall, throws a piton downward. The positions 'y' of the pitons versus 't' during the falling are given in figure. With what speed is the second piton was thrown?

• A. 0 m/s
• B. 7.5 m/s
• C. 17.5 m/s
• D. 27.5 m/s

Answer 1

The correct option is C

17.5 m/s

Let the first piton be dropped from A and the second through B. From the graph, it is evident that B throws the piton 1 second later and they meet at t = 3s

Time of flight for B is 2s and for A is 3s. using $s=ut+\frac{1}{2}a{t}^2$

For A

$\Rightarrow x=0\left(t\right)+\frac{1}{2}(-10)(3{)}^2$

$x=-5\times 9=-45cm\Rightarrow x=-45cm$

For B,

$x-10=u\left(2\right)+\frac{1}{2}(-10)\left(2^2\right)$

$\Rightarrow -45-10=2u-5\times 4$

$\Rightarrow -55=22u-20$

$\Rightarrow 2u=-35$

$\Rightarrow u=\frac{-35}{2}=-17.5m/s$