Question

A ball thrown up vertically returns to the thrower after $6s$. Find (i) velocity (ii) Maximum height it reaches (iii) Position after $4s$

Answer 1

Step 1: Given Data

Total time $T=6s$

$\therefore$Time taken to reach the maximum height $t=\frac{T}{2}=3s$

Final velocity $v=0$

Let the initial velocity be $u$.

Let the acceleration due to gravity be $g=9.8m/s^2$

Let the height be $h$.

Step 2: Calculate the Initial Velocity

According to the first equation of motion,

$v=u+at$

Since the ball, is moving up against gravity, the value of $g$ will be negative.

$\Rightarrow 0=u-gt$

$\Rightarrow u=gt$

$\Rightarrow u=9.8\raisebox{1ex}{$m$}\!\left/ \!\raisebox{-1ex}{$s^2$}\right.\times 3s$

$\Rightarrow u=29.4m/s$

Step 3: Calculate the Maximum Height

According to the third equation of motion,

$v^2=u^2+2aS$

Since the ball, is moving up against gravity, the value of $g$ will be negative.

$\Rightarrow 0=29.4^2-2\times 9.8\times h$

$\Rightarrow 29.4^2=2\times 9.8\times h$

$\Rightarrow h=\frac{29.4\times 29.4}{2\times 9.8}$

$\Rightarrow h=44.1m$

Step 4: Calculate the new Height

Given new time is $4s$.

Since the ball reaches a maximum height at $t=3s$, the ball will move downwards after $4s$.

Therefore, the new time $t\text{'}=4-3=1s$

In this case, the initial velocity will be $u=0$

According to the second equation of motion,

$S=ut+\frac{1}{2}a{t}^2$

Since the ball, is moving down towards gravity, the value of $g$ will be positive.

$\Rightarrow h\text{'}=0+\frac{1}{2}gt{\text{'}}^2$

$\Rightarrow h\text{'}=\frac{1}{2}\times 9.8\raisebox{1ex}{$m$}\!\left/ \!\raisebox{-1ex}{$s^2$}\right.\times 1s^2$

$\Rightarrow h\text{'}=4.9m$ (4.9m below from maximum height)

Hence, (i) velocity is $29.4m/s$ (ii) Maximum height it reaches is $44.1m$ (iii) Position after $4s$ is $4.9m$.