A particle is projected from the ground with a velocity of 25 m / s. After 2 second, it just clears a wall 5 m height. Then angle of projection of particle is [ g =10 m / s 2]

Question

A particle is projected from the ground with a velocity of 25 m / s. After 2 second, it just clears a wall 5 m height. Then angle of projection of particle is [ g =10 m / s 2] (A) 30° (B) 45° (C) 60° (D) 75°

Tardigrade Admin · Accepted Answer

uy=u sin θ y=uy t-(1/2) g t2 or 5=(25 sin θ) × 2-(1/2) × 10 × 22 ∴ sin θ=(1/2) or θ=30°

Q. A particle is projected from the ground with a velocity of $25 m / s$ . After 2 second, it just clears a wall $5 m$ height. Then angle of projection of particle is $[g = 10 m / s^{2}]$

$3 0^{\circ}$

$4 5^{\circ}$

$6 0^{\circ}$

$7 5^{\circ}$

$u_{y} = u sin θ$
$y = u_{y} t - \frac{1}{2} g t^{2}$
or $5 = (25 sin θ) \times 2 - \frac{1}{2} \times 10 \times 2^{2}$
$∴ sin θ = \frac{1}{2}$
or $θ = 3 0^{\circ}$

Q. A particle is projected from the ground with a velocity of 25m/s. After 2 second, it just clears a wall 5m height. Then angle of projection of particle is [g=10m/s2]

30∘

45∘

60∘

75∘