Question

A ball thrown up vertically returns to the thrower after 12 seconds. Find the maximum height it reaches.

Answer 1

Step 1: Given data:

The time for the ball to return is 12 seconds.

Therefore,

⸫ Time taken to reach the maximum height from the ground is = $\frac{12}{2}=6seconds$

Since it takes $6seconds$ for the ball to return. Then,

Step 2: 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}$}\right.$

And

Acceleration, $g=-10m{s}^{-2}$ (Acceleration is negative because the 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: Calculating the maximum height

We know v, a and t

Finding distance using 2^nd equation of motion

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

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

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

= $360-180$

$s=180m$

Thus, the maximum height it reaches is $180m.$