A ball is dropped from a height of 1.5m. how long will it take the ball to reach the ground and what velocity will it reach?

Answer 1

x = x(initial) + v(initial)t + 1/2 at^2
0 = 1.5 + 0 + 1/2 (-10)t^2
1.5 = 5 t^2
t = 0.55 seconds

v = v(initial) + at
v = 0 + (-10)(0.55)
v = -5.5 m/s (i.e. it will reach the ground at 5.5 m/s)

Answer 2

I didn't get anything of what you just wrote here.

Answer 3

s = u + 1/2(a) t^2
1.5 = 0m/s + 1/2 (10 or 9.81) * t^2
1.5 = 1/2 (10 or 9.81~) * t^2
√(1.5/(1/2 (10 or 9.81)))

Answer 4

(Given Information): d = 1.5m, Aav= 9.8 m/s^2, Vi= 0.0 m/s

d= (Vi x t) + (1/2 x Aav x t^2)

Rearrange to get

√(d/1/2 x Aav) = t
√(1.5m / 1/2 x 9.8 m/s^2) = t
0.55s = t