First slide

Projectile motion

Question

A stone A is thrown at an angle of 45^o to the horizontal. Another stone B is thrown with 20 ms^-1 horizontally as shown in figure. They collide in mid-air. If distance BC
is 4n meter, then the value of n is ____.

Difficult

Solution

$\begin{array}{l} v_{A} = u_{A} - g \hat{j} and v_{B} = u_{B} - g t \hat{j} \\ ∴ v_{A B} = v_{A} - v_{B} = u_{A} - u_{B} \\ = 10 \hat{i} - 10 \hat{j} + 20 \hat{i} = 30 \hat{i} - 10 \hat{j} \\ ∴ \tan θ = \frac{1}{3} \end{array}$
It means relative velocity of A with respect to B is directed at $θ$ with vertically downward direction.
For collision, v_AB should be towards initial position of B.
$\begin{array}{l} ∴ \tan θ = \frac{B C}{36} \\ ∴ B C = 36 \times \frac{1}{3} = 12 = 4 n \\ ∴ n = 3 \end{array}$

Get Instant Solutions

When in doubt download our app. Now available Google Play Store- Doubts App

Recieve an sms with download link

Doubts App

OR

SCAN ME

Doubts App

Doubts App