Question

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

Answer 1

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