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Question

$(1)$ Let the speed of particle at position$A$ be $'u'$ and at $B$ be $'v'.$

It took $t_{1}$ time to reach $\mathrm{B}$ from $\math

Vedclass · Accepted Answer

$\frac{1}{2}\, gt_1t_2$

Vedclass · Answer

$(1)$ Let the speed of particle at position$A$ be $'u'$ and at $B$ be $'v'.$

It took $t_{1}$ time to reach $\mathrm{B}$ from $\mathrm{A}$.

Then, using Newton's Equation of motion,

$v=u-g t_{1}$

$(2)$ It took $t_{2}$ time to reach from $\mathrm{B}$ to $\mathrm{A}$ after covering maximum height.

We know, when particle reaches the ground its speed will be $&lsquo;u'.$ (downward)

Again,

$-u=v-g t_{2}$

$(3)$ Using above two equations to get,

$u=\frac{1}{2} g\left(t_{1}+t_{2}\right)$

$(4)$ We know, it took $t_{1}$ time to move from $A$ to $B$

Using Newton's Equation of motion,

$S=u t_{1}-\frac{1}{2} g t_{1}^{2}$

$=\frac{1}{2} g\left(t_{1}+t_{2}\right) t_{1}-\frac{1}{2} g t_{1}^{2}$

$=\frac{1}{2} g t_{1}^{2}+\frac{1}{2} g t_{1} t_{2}-\frac{1}{2} g t_{1}^{2}$

$=\frac{1}{2} g t_{1} t_{2}$

Hence, Height of point $'B'$ from ground is

$\frac{1}{2} g t_{1} t_{2}$

A particle is projected vertically upwards from a point $A$ on the ground. It takes $t_1$ time to reach a point $B$ but it still continues to move up. If it takes $t_2$ time to reach the ground from point $B$ then height of point $B$ from the ground is

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