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

Answer:

1.78 s

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= $= \frac{-b\pm \sqrt{b^2-4\times a \times c}}{2\times a}$

t = $\frac{-7.95 \pm \sqrt{7.95^2-4\times 4.9 \times (-29.8)}}{2\times 9.8}$ = 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