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# Physics Problems: Kinetics and Gravity

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#### Solution Preview

1.)
Height of the building h = 16 m
g = 9.8 m
time of fall through the height of the building t = sqrt(2*h/g)
=> t = sqrt(2*16/9.8) = 1.80 sec

At the time when 1st drop toucues the ground, the fifth one is leaving the roof. Therefore, there
should be three drops in between 1st and 5th. Hence,
t = 1.80 s = 4*T
=> T = interval between two successive drops = 1.80/4 = 0.45 s

It means, at the time when 1st drop landing, the second drop has travelled by 3*T time, 3rd by 2*
T, and 4th by T time.

Hence, the position of 2nd drop from the ground
h2 = 16 - (1/2)*g*(3*T) = 16 - 0.5*9.8*3*0.45 = 9.385 m --Answer

3rd drop,
h3 = 16 - 0.5*9.8*2*0.45 = 11.59 m --Answer

4th drop,
h4 = 16 - 0.5*9.8*0.45 = 13.795 m --Answer

2.)
Initital velocity u = vo (upward)
Final velocity v = ?
displacement s = 0 (coming back to the original point)
acceleration a = g (downward)

Because,
v^2 = u^2 + 2*a*s
=> v^2 = vo^2 + 2*(-g)*0
=> v = - vo
Here, -ve sign shows that the object will have doenward velocity.

Magnitude of v = vo --Proved

For rising,
u = vo (upward)
At the top most point, velocity v = 0
time of rise t = t1 = ?
acceleration a = g (downward)
Because,
v = u + a*t
=> 0 = vo - g*t1
=> t1 (time of rise) = vo/g

Height of the rise h =?
v^2 = u^2 + 2*a*s
=> 0 = vo^2 - 2*g*h
=> h = vo^2/2g

Now, for fall,
u = 0
a = g (downward)
displacement s = rise = h = vo^2/2g

Hence ...

#### Solution Summary

The solution provides step-by-step workings for a number of physics-related calculations. The main concepts covered are kinetics and gravity.

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