From the top of a tower \(60 \mathrm{~m}\) tall, a body is thrown vertically down with a velocity of \(10…

From the top of a tower \(60 \mathrm{~m}\) tall, a body is thrown vertically down with a velocity of \(10 \mathrm{~ms}^{-1}\). At the same time, another body is thrown vertically upward from the ground with a velocity of \(20 \mathrm{~ms}^{-1}\). (a) After how long will the two bodies meet? (b) At what height above the ground do they meet? Take \(g=10\) \(\mathrm{ms}^{-2} .\)
  1. 2 s, 20 m
  2. 4 s, 30 m
  3. 5 s, 20 m
  4. 7 s, 30 m

Solution

(a) Suppose the bodies meet at \(C\) and let \(t\) be the time at which they meet.
\(\begin{array}{l}\text {For body 1: } & s=-h_{1}, u_{1}=-10 \mathrm{~ms}^{-1}, a=-10 \mathrm{~ms}^{-2} \\ \therefore \quad & -h_{1}=-10 t+\frac{1}{2} \times(-10) t^{2}\end{array}\)
which gives \(\quad h_{1}=5 t(t+2)\) __________(1) $\begin{aligned} \text { For body 2: } & \quad s =+h_{2}, u_{2}=+20 \mathrm{~ms}^{-1}, \\ & \quad a =-10 \mathrm{~ms}^{-2} \\ \therefore & \quad h_{2} =20 t-5 t^{2}=5 t(-t+4) \quad \text{(2)} \end{aligned}$ Adding (1) and (2), \(h_{1}+h_{2}=30 t\) or \(60=30 t\) \(\Rightarrow \quad t=2 \mathrm{~s}\).
(b) Using \(t=2 \mathrm{~s}\) in Eq. (2), \(h_{2}=5 \times 2(4-2)=20 \mathrm{~m}\)

Asked in: JEE Mains - Motion In One Dimension - Test 4

Practice more Motion In One Dimension questions on Aicharya