Vertical Motion

Try this problem, useful for Physics Olympiad based on Vertical Motion.

The Problem: Vertical Motion

A boy is standing on top of a tower of height 85m and throws a ball in the vertically upward direction at a certain speed. If 5.25 secs later he hears the ball hitting the ground, then the speed with which the boy threw the ball is ( take g=10m/s2, speed of sound in air=340m/s)
(A)   6m/s
(B)   8m/s
(C)   10m/s
(D)   12m/s

Solution:

Time taken by sound= (\frac{85}{340})=0.25secs
Time taken by the ball= 5.25-0.25=5s
Now,
Let us assume that the ball is thrown upwards with a velocity u. We know, (s=ut+\frac{1}{2}gt^2) where s is the distance covered,u is the initial velocity, g is the acceleration due to gravity and t is the time taken. Here, time taken t=5. Since the ball is thrown upwards, we have $$85=-5u+\frac{1}{2}\times10\times25$$
Hence, u is 8m/s.

Speed of ball thrown down from a height

A boy is standing on top of a tower of height 85m and throws a ball in the vertically upward direction with a certain speed. If 5.25 secs later he hears the ball hitting the ground, then the speed with which the boy threw the ball is ( take g=910m/s², speed of sound in air=340m/s)

6m/s
8m/s
10m/s
12m/s

Solution:

Time taken by sound= 85/340=0.25secs

Time taken by the ball= 5.25-0.25=5 sec

Now,

we write the equation $$ s=ut+1/2ft^2 $$

Here s=-85 and time t =5 secs

Hence,

$$ -85=5u-1/21025$$

$$\Rightarrow u=8m/s$$

Motion under Constant Gravity

Let's discuss a beautiful problem useful for Physics Olympiad based on Motion under Constant Gravity.

The Problem: Motion under Constant Gravity

A person throws vertically up n balls per second with the same velocity. He throws a ball whenever the previous one is at its highest point. The height to which the balls rise is

(a) g/n²

(b) 2gn

(d) 2gn²

Solution:

We know v=u-gt. v is zero at the highest point. Time t taken by one ball to reach maximum height is 1/n.

Hence, we have

u=gt

or, u=g/n……. (i)

now, from the relation v²=u²-2gh. Again, v=0

u²=2gh……. (ii)

Putting the value of u from equation (i) in above relation (ii), we get

h=g/2n².

Motion of an Elevator

Try this problem useful for the Physics Olympiad based on Motion of an Elevator.

The Problem: Motion of an Elevator

An elevator of mass M is accelerated upwards by applying a force F. A mass m initially situated at a height

of 1m above the floor of the elevator is falling freely. It will hit the floor of the elevator after a time equal to

√(2M/F+mg)
√(2M/F-mg)
√2M/F
√2M/(F+Mg)

Discussion:

Acceleration of elevator a_e= F/M ( in upward direction)

Acceleration due to gravity is in downward direction so acceleration of mass a_s= -g

Acceleration of mass with respect to elevator

= a_s-a_e

=(F/M)+g

=(F+Mg)/M

We know,

s=ut+(1/2)at²

From the given problem, we have s=1m

so,

1=(F+Mg)t²/M

t=√2M/(F+Mg)

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