A particle is projected vertically upwards from a point A on the ground. It takes time t1, to reach a point B, but it still continues to move up. It takes further time t2 to reach the ground from point B then height of point B from the ground is: (a) \(\frac{1}{2}g(t_{1}+t_{2})^{2}\) (b) \(gt_{1}t_{2}\) (c) \(\frac{1}{8}g(t_{1}+t_{2})^{2}\) (d) \(\frac{1}{2}gt_{1}t_{2}\)

Given

Total time of particle = (t1 + t2)

To find initial value

Applying 2nd equation of motion

\(0 = u(t_{1}+t_{2})-\frac{g}{2}(t_{1}+t_{2})^{2}\)

We get,

\(\frac{g}{2}(t_{1}+t_{2})= u\)

Since height at t = t1+ t2 = 0

At t = t1,height attained is l

Hence,

\(h = ut_{1}-\frac{g}{2}t_{1}^{2}\) \(h = \frac{g}{2}(t_{1}+t_{2})t_{1}-\frac{g}{2}t_{1}^{2}\) \(h = \frac{g}{2}t_{1}[t_{1}+t_{2}-t_{1}]\)

We get,

\(h = \frac{g}{2}t_{1}t_{2}\)

Therefore, the correct option is (d)

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