Question

A particle is projected from a tower of height 25 m with velocity 20sqrt(2)m//s at 45^@. Find the time when particle strikes with ground. The horizontal distance from the foot of tower where it strikes. Also find the velocity at the time of collision.

Answer 1

Here is your answer:

5 sec

Explanation:

Solution :

u(x) = u(y )=20m/s,ay = −10m/s2

s(y) = uyt+1/2ayt2

−25=20t−1/2×10×t(2)

−25 =20t−12×10×t2

Solving this equation, we get the positive value of ,

t = 5sec

Now apply, v= u+at and s(x) = u(x)t .