Question

Two projectiles are thrown from the same point simultaneously with same velocity 10 ms^-1. One goes straight vertically while other at 60° with the vertical. What will be the distance of separation between the two after 1 second of their throw?

Moderate

Solution

Here for the projectile thrown at 60° with the vertical.
Horizontal velocity = 10 sin 60° = $5 \sqrt{3} {ms}^{- 1}$
vertical velocity = 10 cos 60° = 5 ms^-1
Horizontal separation $Δx = 5 \sqrt{3} \times 1 = 5 \sqrt{3}$
Vertical separation = h₁ - h₂
$\begin{array}{l} Δy = (10 t - \frac{1}{2} g t^{2}) - [5 t - \frac{1}{2} g t^{2}] \\ = (10 \times 1 - 5 \times 1) = 5 m \end{array}$
$∴$ displacement $| r | = \sqrt{(5 \sqrt{3})^{2} + 5^{2}} = 10 m$

Get Instant Solutions

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

Recieve an sms with download link

SCAN ME