First slide

Projection Under uniform Acceleration

Question

A particle is projected with a velocity 40 m/s at an angle 30⁰with the horizontal. The magnitude of change in velocity when it is at a height of 15 m from the ground is

Moderate

A

10m/s
B

20m/s
C

5 m/s
D

30 m/s

Solution

There is no change in horizontal velocity
Vertical velocity at a height h
$\begin{array}{l} v_{y}^{2} = (usin θ)^{2} - 2 gy \\ v_{y}^{2} = {(40 \frac{1}{2})}^{2} - 2 (10) 15 \\ v_{y}^{2} = 100 \\ v_{y} = 10 m / s \end{array}$
Change in velocity
$\begin{array}{l} u \sin θ - 10 \\ 40 (0.5) - 10 \\ 20 - 10 = 10 m / s \end{array}$

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