Question

Two balls are projected vertically upward with the same speed of $35m/s$ in a $3s$ interval, find the height in $\mathrm{meters}$ at which both balls collide.

• A. $50$
• B. $30$
• C. $80$
• D. $100$

Answer 1

The correct option is A

$50$

Step 1: Given Data

The initial velocity of both the balls $u_1=u_2=35m/s$

Time taken by the first ball $=t$

Time taken by the second ball $=t-3$

Let the distance traveled by both the balls be $S_1$ and $S_2$.

Let the acceleration due to gravity be $g=-10m/s^2$ since the balls are moving upwards.

Step 2: Formula Used

Newtown's second equation of motion is given as,

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

Step 3: Calculate the Time Period

Since balls collide, their height must be equal.

$$\begin{gathered} \therefore S_1=S_2 \\ \Rightarrow 35t-\frac{1}{2}g{t}^2=35\left(t-3\right)-\frac{1}{2}g{\left(t-3\right)}^2 \\ \Rightarrow 35t-\frac{1}{2}g{t}^2=35t-35\times 3-\frac{1}{2}g\left(t^2-6t+9\right) \\ \Rightarrow -\frac{1}{2}g{t}^2=-35\times 3-\frac{1}{2}g{t}^2+3gt-\frac{9}{2}g \\ \Rightarrow 0=-35\times 3+3\times 10t-\frac{3\times 3}{2}\times 10 \\ \Rightarrow 0=-35+10t-15 \\ \Rightarrow 10t=50 \\ \Rightarrow t=5s \end{gathered}$$

Step 4: Calculate the height

Therefore, the height can be calculated as,

$$\begin{gathered} h=35\times 5-\frac{1}{2}\times 10\times 5^2 \\ =175-125 \\ =50m \end{gathered}$$

Hence, the correct answer is option (A).