Question

A young girl standing on a bridge throws a stone with an initial velocity of 12 m/s at a downward angle of 45° to the horizontal, in an attempt to hit a block of wood floating in the river below. If the stone is thrown from a height of 20 m and it just reaches the water when the block is 13 m from the bridge, does the stone hit the block?

Answer 1

Solution.

$v_0=12 m/s;$

$\theta=45^o;$

$v_{x0}=v_ocos45^o=8.5 m/s;$

$v_{y0}=-v_osin45^o=-8.5 m/s;$

$y=-20m;$

$x_b=13m; (x_0=y_0=0);$

$v_y^2=v_{y0}^2-2gh;$

$v_y=\sqrt{144-2\sdot9.81\sdot(-20)}=-22m/s;$

$v_y=v_{0y}-gt\implies t=\dfrac{v_{y0}-v_y}{g};$

$t=\dfrac{-8.5-(-22)}{9.81}=1.4s;$

$x_{max}=v_{0x}t=8.5\sdot1.4=12m;$

Answer: The girl's throw falls short by a meter( the block is at 13m).