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# 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.

Given data:

$$g = 10 m/s^2$$

Total time T = 6 sec

$$T_a = T_d = 3 sec$$

(a) Final velocity at maximum height v = 0

We know that, $$v = u – gt_a \$$
$$\ u = v + gt_a \$$
$$\Rightarrow \ u = 0 + 10 × 3$$

$$\Rightarrow u = 30 m/s$$

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

(b) Now, $$s = ut_a - \frac{1}{2}g(t_a)^2 \$$
$$\Rightarrow \ s = 30 × 3 - \frac{1}{2} × 3^2 \$$
$$\Rightarrow \ s = 45 m$$

The maximum height stone reaches is 45 m.

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

Distance travelled in another 1 sec = s’

So, $$s = ut_a - \frac{1}{2}g(t_a)^2 \$$
$$\Rightarrow \ s = 0 + 10 × 1 × 1 \$$
$$\Rightarrow \ s = 5 m$$

The distance travelled in another 1 sec = 5 m.

$$\therefore$$ in 4 sec, the position of point p (45 – 5)

= 40 m from the ground.