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Standard XII

Physics

Vector Component

A stone is pr...

Question

A stone is projected from the ground with a velocity of 25 m/s. Two seconds later, it just clears a pole of 5m height. The angle of projection is

45°

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30°

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60°

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75°

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Solution

The correct option is B
30°

$u_{y}$ = 25 sin $θ$

After 2 second, the height reached is 5m.

h = $u_{y}$ t - $\frac{1}{2}$ g $t^{2}$ ,

Substituting, we get,

$θ$ = $30^{\circ}$

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A stone is projected from the ground with a velocity of 25 m/s. Two seconds later, it just clears a pole of 5m height. The angle of projection is

The correct option is B 30° uy = 25 sin θ After 2 second, the height reached is 5m. h = uyt - 12 g t2 , Substituting, we get, θ = 30∘

The correct option is B
30°

$u_{y}$ = 25 sin $θ$

After 2 second, the height reached is 5m.

h = $u_{y}$ t - $\frac{1}{2}$ g $t^{2}$ ,

Substituting, we get,

$θ$ = $30^{\circ}$