A ball is projected vertically upward from the ground. It crosses a point 20 m above the point of projection twice, and the time interval between the two crossings is 2 s. Neglect air resistance and take g = 10 m/ $s^2.$ The initial speed of projection is:
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Question

A ball is projected vertically upward from the ground. It crosses a point 20 m above the point of projection twice, and the time interval between the two crossings is 2 s. Neglect air resistance and take g = 10 m/ $s^2.$ The initial speed of projection is:

Options

G1 Physics · Accepted Answer

At height 20 m: 20 = ut - 5 t 2 , t^2, t 2 , or 5 t 2 t^2 t 2 - ut + 20 = 0. If t1 and t2 are the two roots, t2 - t1 = sqrt( u 2 u^2 u 2 - 400)/5. Given t2 - t1 = 2, so sqrt( u 2 u^2 u 2 - 400) = 10. Thus u 2 u^2 u 2 = 500 and u = 10sqrt(5) m/s. Hence, option B is correct.

A ball is projected vertically upward from the ground. It crosses a point 20 m above the point of projection twice, and the time interval between the two crossings is 2 s. Neglect air resistance and take g = 10 m/ s 2 . s^2. s 2 . The initial speed of projection is: