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A ball is thrown upward from the ground with an initial speed of $v$. At the same instant, another ball is dropped from a building of height $20 \mathrm{~m}$. If the balls are at the same height after $0.8 \mathrm{~s}$, then the magnitude of $v$ is
(Take, acceleration due to gravity $=10 \mathrm{~ms}^{-2}$ )
Options:
(Take, acceleration due to gravity $=10 \mathrm{~ms}^{-2}$ )
Solution:
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Verified Answer
The correct answer is:
$25 \mathrm{~ms}^{-1}$
The given situation is shown below
Both balls meet at point $C$ after $0.8 \mathrm{~s}$ i.e
Let
$$
t_1=t_2=0.85
$$
$$
\therefore \quad B C=20-h
$$
For the first ball,
$$
\begin{aligned}
h & =v t_1-\frac{1}{2} g t_1^2 \\
\Rightarrow h & =v \times 0.8-\frac{1}{2} \times 10 \times 0.8^2
\end{aligned}
$$
For the second ball,
$$
\begin{aligned}
20-h & =0+\frac{1}{2} g t_2^2 \\
20-h & =\frac{1}{2} \times 10 \times(0.8)^2 \\
\Rightarrow \quad 20-h & =3.2 \\
\Rightarrow \quad h & =20-3.2 \\
h & =16.8 \mathrm{~m}
\end{aligned}
$$
Putting the value of $h$ in Eq. (i), we get
$$
\begin{array}{rlrl}
& 16.8 & =v \times 0.8-\frac{1}{2} \times 10 \times 0.8^2 \\
\Rightarrow \quad 16.8 & =0.8 v-3.2 \\
\Rightarrow \quad 20 & =0.8 v \\
\Rightarrow \quad & v & =\frac{20}{0.8}=25 \mathrm{~ms}^{-1}
\end{array}
$$
Both balls meet at point $C$ after $0.8 \mathrm{~s}$ i.e
Let
$$
t_1=t_2=0.85
$$
$$
\therefore \quad B C=20-h
$$
For the first ball,
$$
\begin{aligned}
h & =v t_1-\frac{1}{2} g t_1^2 \\
\Rightarrow h & =v \times 0.8-\frac{1}{2} \times 10 \times 0.8^2
\end{aligned}
$$
For the second ball,
$$
\begin{aligned}
20-h & =0+\frac{1}{2} g t_2^2 \\
20-h & =\frac{1}{2} \times 10 \times(0.8)^2 \\
\Rightarrow \quad 20-h & =3.2 \\
\Rightarrow \quad h & =20-3.2 \\
h & =16.8 \mathrm{~m}
\end{aligned}
$$
Putting the value of $h$ in Eq. (i), we get
$$
\begin{array}{rlrl}
& 16.8 & =v \times 0.8-\frac{1}{2} \times 10 \times 0.8^2 \\
\Rightarrow \quad 16.8 & =0.8 v-3.2 \\
\Rightarrow \quad 20 & =0.8 v \\
\Rightarrow \quad & v & =\frac{20}{0.8}=25 \mathrm{~ms}^{-1}
\end{array}
$$
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