from the top of a multi storey building, 39.2 m tall a boy projects a stone vertically upward with u=9.8 m/s such that it finally drop to ground.when will the stone reach the ground? when will it passs through the point of projection? what will be its velocity before striking the ground?
Answers
a) Time taken to reach the ground
The height reached by the projectile from point of projection is :
v² = u² - 2gs
g = 9.8
U = 9.8
v = 0
0 = 96. 04 - 19.6S
19.6S = 96.04
S = 96.04/19.6
S = 4.9m
The time taken to reach this height is :
t = u/g
= 9.8/9.8 = 1
Total height from the ground : 4.9 + 39.2 = 44.1
It takes 2 secs to fall back to the point of projection.
The final velocity would be:
V = u + gt
V = 0 + 9.8
V = 9.8
This is the initial velocity with which it falls from the point of projection to the ground.
Now using the formulae below, we can get time it takes to fall from point of projection.
S = 39.2
S = ut + 0.5gt²
39.2 = 9.8t + 4.9t²
Dividing through by 4.9
8 = 2t + t²
t² + 2t - 8 = 0
Solving by quadratic method we have:
The roots are : (-2, 4)
t² + 4t - 2t - 8 = 0
t(t + 4) - 2(t + 4) = 0
(t+4)(t - 2) = 0
t = - 4 or 2
So we take 2 since it is positive.
The total time taken to reach the ground is thus :
2 + 2 = 4 seconds
Answer : 4 seconds.
b) Time taken to pass through point of projection is :
From the working above, the answer is 2
c)Velocity before striking the ground.
We will use the formulae below:
V² = U² + 2gs
U = 9.8
S = 39.2
g = 9.8
V² = 96.04 + 768.32
V = √864.36 = 29.4
= 29.4m/s
Answer:
Amswer is given above.
