First slide

Projection Under uniform Acceleration

Question

A particle is projected upwards with a velocity of 100 m/sec at an angle of 60° with the vertical. Find the time when the particle will move perpendicular to its initial direction, taking g = 10 m/sec²:

Moderate

A

10 second
B

20 second
C

5 second
D

10 $\sqrt{3}$ second

Solution

$Method I : \vec{u} = 100 \cos 30 ° \hat{i} + 100 \sin 30 ° \hat{j} = (50 \sqrt{3} \hat{i} + 50 \hat{j})$
$v = \vec{u} + \vec{g} t = (50 \sqrt{3} \hat{i} + 50 \hat{j}) + (- 10 j) t = 50 \sqrt{3} \hat{i} + (50 - 10 t) \hat{j}$
$if \vec{u} ⊥ \vec{v}, then \vec{u} . \vec{v} = 0 so {(50 \sqrt{3})}^{2} + 50 (50 - 10 t) = 0$
$\Rightarrow t = 20 second$

$Method II : By v = v + at = u + gt$
From right angled triangle
$\frac{u}{gt} = \cos 60 ° \Rightarrow t = 20 s$

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