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.