Question

A solid steel ball is thrown directly downward, with an initial speed of 7.95 m/s, from the top of a building at a height of 29.8 m. How much time (in s) does it take before striking the ground?

Answer 1

1.78 s

Step-by-step explanation:

Initial speed of the ball = u = 7.95 m/s and is vertically downwards.

Acceleration due to gravity = g = 9.8 m/s/s , vertically downwards.

Height of the building h = 29.8 m (traversed downwards by the steel ball).

h = u t + 1/2 g t²

29.8 = 7.95 t + 0.5 (9.8) t²

⇒ 4.9 t² +7.95 t - 29.8 = 0

Using the quadratic formula , solve for t.

t=
`= (-b\pm √(b^2-4* a * c))/(2* a)`

t =
`(-7.95 \pm √(7.95^2-4* 4.9 * (-29.8)))/(2* 9.8)` = 1.78 s, -3.4 s

Since time does not have a negative value, time taken by the stone to reach the ground = t = 1.78 s