Question

# A ball is thrown horizontally from the top of a tower $$40$$m high. The ball strikes the ground at a point $$80$$m from the bottom of the tower. Find the angle that the velocity vector makes with the horizontal just before the ball hits the ground.

A
45m/s
B
90m/s
C
37m/s
D
53m/s

Solution

## The correct option is D $$45 m/s$$initial vertical velocity v is 0 distance covered in the vertical direction is 40 m $$s=ut+\dfrac { 1 }{ 2 } { at }^{ 2 }$$$$40=\dfrac { 1 }{ 2 } \times 10\times { t }^{ 2 }$$$$=t=2\sqrt { 2 } seconds$$horizontal velocity is u $$ut=90=u={ 90 }{ 2\sqrt { 2 } }={ 45 }\dfrac { m }{ s }$$Physics

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