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 4s.

Asked by Vishal kumar | 1 year ago |  198

##### Solution :-

Given data:

g = 10m/s2

Total time T = 6sec

Ta = Td = 3sec

(a) Final velocity at maximum height v = 0

From first equation of motion:-

v = u - gta

u = v + gta

= 0 + 10 x 3

= 30m/s

The velocity with which the stone was thrown up is 30m/s.

(b) From second equation of motion

$$s = ut_a - \frac{1}{2}g(t_a)^2$$

$$= 30 \times 3 - (1/2) \times 10 \times (3)^2$$

$$= 90-45 = 45m$$

The maximum height stone reaches is 45 m.

(c) In 3sec, it reaches the maximum height.

Distance travelled in another 1sec = s’

$$s = ut_a - \frac{1}{2}g(t_a)^2$$

$$s = 0 + 10 \times 1 \times 1$$

s = 5m.

The distance travelled in another 1sec = 5m.

Therefore in 4sec, the position of point p (45 – 5)

= 40m from the ground.

Answered by Shivani Kumari | 1 year ago

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