Chapter - 3 Gravitation

Q&A -Ask Doubts and Get Answers

Q
Gravitation

Question:

A ball thrown up vertically returns to the thrower after 6 s. Find (a) the velocity with which it was thrown up, (b) the maximum height it reaches, and (c) its position after 4 s.

Answer:

(a) To find: The velocity with which it was thrown up.

Time of ascent is equal to the time of descent. The ball takes a total of 6s for its upward and downward journey.

Hence, time taken for the upward journey,

t = \frac{6}{2} = 3\ s

At maximum height final velocity v = 0

Using the first equation of motion:

v = u + at

0 = u -9.8\times 3

u = 29.4\ ms^{-1}

The velocity with which the stone was thrown up is 29..4 m/s.

(b) To find: The maximum height ball reaches.

Using the second equation of motion

s = ut + \frac{1}{2}at^2

s = 29.4 \times 3 + \frac{1}{2} \times -9.8 \times (3)^2

s = 88.2 - 4.9 \times 9

s = 44.1\ m

The maximum height ball reaches is 44.1 m.

(c) To find: Ball's position after 4 sec.

Ball attains the maximum height after 3s. After attaining this height, it will start falling downwards.

In this case, Initial velocity, u = 0

Position of the ball after 4 s of the throw is given by the distance travelled by it during its downward journey in 4 s - 3 s = 1 s

Distance travelled in another 1sec = s'

s' = ut + \frac{1}{2}at^2

s' = 0 \times 1 + \frac{1}{2} \times 9.8 \times (1)^2

s' = 4.9\ m

The distance travelled in another 1 sec = 4.9 m.

Therefore in 4 sec, the position of ball =  (44.1 - 4.9) = 39.2 m from the ground.


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