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Question

A ball thrown up vertically returns to the thrower after 12 second. Find
(i) velocity with which it was thrown up.
(ii) the maximum height it reaches.
(iii) its position after 4 s

(Take g = 10 m s–2)


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Solution

Step 1: Given data:

Time for the ball to return is 12 seconds.

Therefore,

Time taken to reach the maximum height = 122

=6seconds

Since it takes 6seconds for the ball from the ground to maximum height. Then,

Step 2: (i) Finding Initial Velocity

Let Initial velocity = u

At the highest point of motion, velocity becomes 0

Therefore

Final velocity = v = 0 ms

Acceleration = g=-10ms-2 (Acceleration is negative because ball is thrown upwards)

Time = 6seconds

We know v, a and t

Finding u by 1st equation of motion

v=u+at

0=u+(-10)(6)

0=u60

u=60ms-1

Thus, the initial velocity is60ms-1

Step 3: (ii) Calculating the maximum height

Calculating maximum height

We know v, a and t

Finding distance using 2nd equation of motion

s = ut+12at2

= 60×6+12-1062

= 360-12×10×36

= 360-180

s = 180 m

Thus, the maximum height it reaches is 180m.

Step 4: (iii) Find the position after 4 s.

The ball reaches the highest point in 6 seconds.

At 4 seconds, it will be moving upward.

Initial velocity = 60 ms

Time = 4 seconds

Acceleration = g=-10 ms

We need to find s

We know u, a and t

Finding using 2nd equation of motion.

s=ut+12at2

=60×4+12×-10×42

=240-80

s=160m

Thus, the position of the ball after 4 seconds is 160 m.


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