A body is projected horizontally from the top of a tall tower with a velocity of 30ms^(-1). At time t_(1), its horizontal and vertical components of the velocity are equal and at time t_(2), its horizontal and vertical displacements are equal. Then t_(2)-t_(1) is (take, g=10ms^(-2))

A body is projected horizontally from the top of a tall tower with a v

1.	A body is projected horizontally from the top of a tall tower with a velocity of 30ms^(-1). At time t_(1), its horizontal and vertical components of the velocity are equal and at time t_(2), its horizontal and vertical displacements are equal. Then t_(2)-t_(1) is (take, g=10ms^(-2))
Answer» 1 s 1.5 s 2 s 3 s Solution :As PER FIRST condition, `30=10t_(1)` `t_(1)=3s` As per SECOND condition, `30t_(2)=1/2xx10t_(2)^(2)=t_(2)=6s` `therefore" "t_(2)-t_(1)=3s`

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A body is projected horizontally from the top of a tall tower with a velocity of 30ms^(-1). At time t_(1), its horizontal and vertical components of the velocity are equal and at time t_(2), its horizontal and vertical displacements are equal. Then t_(2)-t_(1) is (take, g=10ms^(-2))

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