Aball is thrown directly downward with an initial speed of 8.50 m/s, from a height of 29.7 m. after what time interval does it strike the ground?

Using the following formula for linear-motion, the missing variable can be solved:

s = Vi * t + 1/2 (a * t^2) 

Where: s = displacement = 29.7 m
Vi = initial velocity = 8.5 m/s
a = acceleration = 9.8
t = time = ?

Substituting:

29.7 = 8.5t + 1/2 (9.8*t^2)
29.7 = 8.5t + 4.9t^2

Dividing both sides by 4.9:

6.06 = 1.73t + t^2
t^2 + 1.73t - 6.06 = 0
(t - 1.74)(t + 3.48) = 0
t = 1.74s

From the above values, the correct answer is 1.74 seconds.

At maximum height, kinetic energy is converted in potential energy  formula: at max height, ke = ½ * 0.27kg * (16m/s) ² = 34.56 j  so the remaining pe = (41.70 – 34.56)= 7.14 j = 0.27 kg * 9.8m/s² * h  = (9.8/7.14 * 0.27)so max height h = 5.08 or 5 m