Question 49: Motion in 1D

Question

Two balls A and B are placed at the top of 180 m tall tower. Ball A is released from the top at t = 0 s. Ball B is thrown vertically down with an initial velocity 'u' at t = 2 s. After a certain time, both balls meet 100 m above the ground. Find the value of 'u' in ms
$$-$$1
. [use g = 10 ms
$$-$$2
] :

Accepted Answer

As the meeting point lies $100 \mathrm{~m}$ above ground, displacement of ball will be $80 \mathrm{~m}$.
For ball $A$
$$
\begin{aligned}
&u=0, \mathrm{~S}=80 \mathrm{~m}, a =+\mathrm{g}=+10 \mathrm{~m} / \mathrm{s}^2 	ext {, time }=t_1 \\
&\Rightarrow S =u t+\frac{1}{2} a t^2 \\
&\Rightarrow 80 =0+\frac{1}{2} 	imes 10 	imes t_1{ }^2 \\
&\Rightarrow \frac{160}{10} =t_1{ }^2 \\
&\Rightarrow \quad t_1 =4 \mathrm{~s}
\end{aligned}
$$
As ball $B$ is thrown after 2 seconds after release of $A$. Thus, time available for ball $B$ is 2 seconds to cover a distance of $80 \mathrm{~m}$.
Let speed be ' $u$ ' $\mathrm{m} / \mathrm{s}, t_2=4-2=2 \mathrm{~s}, \mathrm{~S}=80 \mathrm{~m}$, $a=+g=+10 \mathrm{~m} / \mathrm{s}^2$
$$
	herefore  80  =u 	imes 2+\frac{1}{2} 	imes 10 	imes(2)^2 $$
$$\Rightarrow 80  =u+20 $$
$$\Rightarrow  2 u  =60 $$
$$\Rightarrow u  =30 \mathrm{~m} / \mathrm{s}$$

Answer

10

Answer

15

Answer

20

Answer

30

Motion in 1D

Two balls A and B are placed at the top of 180 m tall tower. Ball A is released from the top at t = 0 s. Ball B is thrown vertically down with an initial velocity 'u' at t = 2 s. After a certain time, both balls meet 100 m above the ground. Find the value of 'u' in ms $-1 . [use g = 10 ms -$2 ] :

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