A body projected vertically upward from the ground is at same height h from the ground at two instant of time t1 and t2 (both measured from instant of projection). Then velocity of projection

Question

A body projected vertically upward from the ground is at same height h from the ground at two instant of time t1 and t2 (both measured from instant of projection). Then velocity of projection (A) (g/2)(t1+t2) (B) g(t, t2) (C) g(t1+t2) (D) (g/4)(t1+t2)

Tardigrade Admin · Accepted Answer

t1+t2=(2 v/g) u=(g/2)(t1+t2) h=u t1-(1/2) g t12=(g/2) t1 t2

Q. A body projected vertically upward from the ground is at same height $h$ from the ground at two instant of time $t_{1}$ and $t_{2}$ (both measured from instant of projection). Then velocity of projection

$\frac{g}{2} (t_{1} + t_{2})$

$g (t, t_{2})$

$g (t_{1} + t_{2})$

$\frac{g}{4} (t_{1} + t_{2})$

$t_{1} + t_{2} = \frac{2 v}{g}$
$u = \frac{g}{2} (t_{1} + t_{2})$
$h = u t_{1} - \frac{1}{2} g t_{1}^{2} = \frac{g}{2} t_{1} t_{2}$

Q. A body projected vertically upward from the ground is at same height h from the ground at two instant of time t1​ and t2​ (both measured from instant of projection). Then velocity of projection

2g​(t1​+t2​)

g(t,t2​)

g(t1​+t2​)

4g​(t1​+t2​)