Question

A particle projected from a tower of height $25m$ with velocity $20\sqrt{2}m/s$ at ${45}^o$. Find the time when particlle strikes with ground. The horizonatl distance from the foot of tower where it strikes

Answer 1

From A to B

s = $ut+\frac{1}{2}a{t}^2$

-25 = $20t-5t^2$

solving above equation

we get

$t^2-4t-5$ = 0

t = 5 sec

$BC=v_x\times t$

= $20\times 5$

= 100m