A particle is projected from the earth's surface with a velocity of 50 m s- 1 at an angle θ with the horizontal. After 2 s it just clears a wall 5 m high. What is the value of 55 sinθ ? ( g=10 m s- 2 )

Question

A particle is projected from the earth's surface with a velocity of 50 m s- 1 at an angle θ with the horizontal. After 2 s it just clears a wall 5 m high. What is the value of 55 sinθ ? ( g=10 m s- 2 ) (A)  (B)  (C)  (D)

Tardigrade Admin · Accepted Answer

<img class="img-fluid question-image" alt="Solution" src="https://cdn.tardigrade.in/q/nta/p-mqrb4bodtd2d.png" /> Using equation of motion in y -dir s=u t+(1/2) a t2 Height after 2 s h=5=(50 sin θ) t-(1/2) g(t)2 5=(50 sin θ)(2)-(1/2) ⋅ 10 ⋅(2)2 (25/100)= sin θ sin θ=(1/4) 55 sin θ=(55/4)=13.75

Q. A particle is projected from the earth's surface with a velocity of $50 m s^{- 1}$ at an angle $θ$ with the horizontal. After $2 s$ it just clears a wall $5 m$ high. What is the value of $55 s in θ$ ? ( $g = 10 m s^{- 2}$ )

Using equation of motion in $y$ -dir $s = u t + \frac{1}{2} a t^{2}$
Height after $2 s$
$h = 5 = (50 sin θ) t - \frac{1}{2} g (t)^{2}$
$5 = (50 sin θ) (2) - \frac{1}{2} \cdot 10 \cdot (2)^{2}$
$\frac{25}{100} = sin θ$
$sin θ = \frac{1}{4}$
$55 sin θ = \frac{55}{4} = 13.75$

Q. A particle is projected from the earth's surface with a velocity of 50ms−1 at an angle θ with the horizontal. After 2s it just clears a wall 5m high. What is the value of 55sinθ ? ( g=10ms−2 )

Using equation of motion in y -dir s=ut+21​at2 Height after 2s h=5=(50sinθ)t−21​g(t)2 5=(50sinθ)(2)−21​⋅10⋅(2)2 10025​=sinθ sinθ=41​ 55sinθ=455​=13.75

Q. A particle is projected from the earth's surface with a velocity of $50 m s^{- 1}$ at an angle $θ$ with the horizontal. After $2 s$ it just clears a wall $5 m$ high. What is the value of $55 s in θ$ ? ( $g = 10 m s^{- 2}$ )

Using equation of motion in $y$ -dir $s = u t + \frac{1}{2} a t^{2}$
Height after $2 s$
$h = 5 = (50 sin θ) t - \frac{1}{2} g (t)^{2}$
$5 = (50 sin θ) (2) - \frac{1}{2} \cdot 10 \cdot (2)^{2}$
$\frac{25}{100} = sin θ$
$sin θ = \frac{1}{4}$
$55 sin θ = \frac{55}{4} = 13.75$