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