Question

A particle is projected from ground with velocity $20\sqrt{2}m/sat{45}^{\circ }.$ At what time particle is at height $15m$ from ground? $(g=10m/s^2)$

Answer 1

Given : $u=20\sqrt{2}m/s$ $a_y=-g=-10m/s^2$

$\therefore$ $u_x=ucos{45}^o=20m/s$ and $u_y=usin{45}^o=20m/s$

y direction : $S_y=u_{y}t+\frac{1}{2}{a}_y{t}^2$ where $S_y=15$ m

$\therefore$ $15=20t-\frac{10\times t^2}{2}$

OR $t^2-4t+3=0$

$⟹$ $t=1s$ and $3s$

Hence the particle is at height 15 m from ground after $1$ s and $3$ s