# Physics Problems: Kinetics and Gravity

Please see attached file for questions.

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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.