First slide

Projection Under uniform Acceleration

Question

Two particles ,4 and B are shot from the same height at t = 0 in opposite directions with horizontal velocities 3 m/s and 4 m/s respectively. If they are subjected to the same vertical acceleration due to gravity(g = 9.8m/s²)'the distance between them when their velocity vectors become mutually perpendicular, is :

Moderate

A

1.059 m
B

1.412 m
C

2.474 m
D

9.8 m

Solution

${\vec{v}}_{A} = 4 i - gt \hat{j} and {\vec{v}}_{A} = - 3 \hat{i} - gt \hat{j}$
If ${\vec{v}}_{A} ⊥ {\vec{v}}_{B}$ then, ${\vec{v}}_{A} . {\vec{v}}_{B} = 0$
-12+ g² t² = 0
$t = \frac{\sqrt{12}}{g}$
Separation between particles = (4 + 3) (t) = $\frac{7 \sqrt{12}}{9.8}$ = 2.474 m

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