Question

a body is dropped from a height of 176.4m. find the distance covered by it in the last second of its motion. with what velocity will the body strike the ground?(g=9.8m/s^2)​

Answer 1

Answer:

V = 58.8 m/s and distance in one second =  53.9 m

Explanation:

Body is dropped from h = s = 176.4 m and with a = g = 9.8 m/s², u = 0

v² = u² + 2*a*s

   = 2*9.8*176.4

solve for v, we get

v  = 58.8 m/s

Now, v = u + a*t

58.8 = 9.8 * t

t = 6 sec

distance traveled in that one second is s = 0.5 * 9.8 * (6² - 5²) = 53.9 m