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)

Answer 1

Step 1: Given data:

Time for the ball to return is 12 seconds.

Therefore,

⸫ Time taken to reach the maximum height = $\frac{12}{2}$

=$6seconds$

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 $\raisebox{1ex}{$\mathrm{m}$}\!\left/ \!\raisebox{-1ex}{$\mathrm{s}$}\right.$

Acceleration = $g=-10m{s}^{-2}$ (Acceleration is negative because ball is thrown upwards)

Time = $6seconds$

We know v, a and t

Finding u by 1^st equation of motion

$v=u+at$

$0=u+(-10)\left(6\right)$

$0=u–60$

$u=60m{s}^{-1}$

Thus, the initial velocity is$60m{s}^{-1}$

Step 3: (ii) Calculating the maximum height

Calculating maximum height

We know v, a and t

Finding distance using 2^nd equation of motion

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

= $60\times 6+\frac{1}{2}\left(-10\right){\left(6\right)}^2$

= $360-\frac{1}{2}\times 10\times 36$

= $360-180$

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 $\raisebox{1ex}{$\mathrm{m}$}\!\left/ \!\raisebox{-1ex}{$\mathrm{s}$}\right.$

Time = 4 seconds

Acceleration = $g=-10$ $\raisebox{1ex}{$\mathrm{m}$}\!\left/ \!\raisebox{-1ex}{$\mathrm{s}$}\right.$

We need to find s

We know u, a and t

Finding using 2^nd equation of motion.

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

$=60\times 4+\frac{1}{2}\times \left(-10\right)\times {\left(4\right)}^2$

$=240-80$

$s=160m$

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