Answer:
2.79s
Explanation:
Suppose the distance from the throwing point to the ground s = 100m. And suppose g = 10m/s2.
We have the following equation of motion for s in term of t
[tex] s(t) = v_0t + gt^2/2[/tex]
where v0 = 8m/s is the inital speed of the rock, and t is the time it takes to cover a distance of s = 100m
[tex] 100 = 8t + 10t^2[/tex]
[tex]10t^2 + 8t - 100 = 0[/tex]
[tex]t = \frac{-b \pm \sqrt{b^2 - 4ac}}{2a}[/tex]
[tex]t = \frac{-8 \pm \sqrt{8^2 - 4*10*(-100)}}{2*10}[/tex]
[tex]t = \frac{-8 \pm 63.75}{20}[/tex]
t = 2.79s or t = -3.587s
Since t can be positive we will pick t = 2.79s
Time taken by stone to reach the ground is 10.28 seconds.
Given that:
Height of tower = 600 m (Assume)
Speed of throw = 8 m/s
Find:
Time taken by stone to reach the ground
Computation:
s = ut + (1/2)(g)(t²)
600 = (8)(t) + (1/2)(9.8)(t²)
600 = 8t + 4.9t²
4.9t² + 8t - 600
So,
t = 10.28 seconds
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