First slide

Instantaneous acceleration

Question

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. If it takes further t₂ time to reach the ground from point B, then height of point B from the ground is :

Easy

A

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

gt₁t₂
C

$\frac{1}{8} g {(t_{1} + t_{2})}^{2}$
D

$\frac{1}{2} {gt}_{1} t_{2}$

Solution

If after t₁ and t₂ sec, it is at the same height, then time of flight
$T = \frac{2 U}{g} = (t_{1} + t_{2}) U = \frac{g}{2} (t_{1} + t_{2})$
At t₁, displacement ‘h’
$h = - {Ut}_{1} + \frac{1}{2} {gt}_{1}^{2}$
Substituting value of ‘U’ $h = \frac{1}{2} g t_{1} t_{2}$

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