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

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 - 5t^2, or 5t^2 - ut + 20 = 0. If t1 and t2 are the two roots, t2 - t1 = sqrt(u^2 - 400)/5. Given t2 - t1 = 2, so sqrt(u^2 - 400) = 10. Thus 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. The initial speed of projection is: