Question

a body is thrown down from a tower of height 192m with initial velocity of 2m/s^2. Find the time taken to hit the ground and velocity just before it hits the ground​

Answer 1

Answer:

Explanation:

v²= u²+2gs

v² =  2² + 2*10*192

v² = 4 + 3840 = 3844

v = √3844 = 62 m/s

now

v = u + at

v - u = at

v - u/a = t

(62 - 2)/10 = t

t = 60/10

t = 6s

Answer 2

apply the formula

= v^2=u^2+2gs