First slide

Projection Under uniform Acceleration

Question

A particle is projected with a velocity 40 m/s at an angle $30^{o}$ with the horizontal. Find the change in velocity when it is at a height of 15 m from the ground, (g = 10 $m / s^{2}$ )

Moderate

A

10 m/s
B

20 m/s
C

5 m/s
D

$20 \sqrt{3} m / s$

Solution

Horizontal velocity remains constant
Vertical velocity changes
Initial vertical velocity $usinθ = 40 \sin 30^{0}$
= 20 m/s
At height 15 m its vertical velocity
${u_{y}}^{2} = 20^{2} - 2 g (15)$
= 400 – 300
${u_{y}}^{2} = 20^{2} - 2 g (15)$
Change in velocity = 20 – 10 = 10 m/s.

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