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?
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).
Comments
Leave a comment