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= [tex]= \frac{-b\pm \sqrt{b^2-4\times a \times c}}{2\times a}[/tex]
t = [tex]\frac{-7.95 \pm \sqrt{7.95^2-4\times 4.9 \times (-29.8)}}{2\times 9.8}[/tex] = 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
