Question

Projectile is projected speed
of 40ms-1 at angle 30 degree to horizontal.magnitude of change in velocity of projectile till t=3s after launching is
a)10ms-1 b)30ms-1 c)20ms-1 d)20√3ms-1​

Answer 1

Given :

Initial speed of projectile = 40 m/s

Angle of projection = 30°

To Find :

Magnitude of change in velocity of projectile till t = 3 after launching.

Solution :

❖ Initial speed of projectile :

$\sf:\implies\:\vec{u}=u\cos\theta\hat{i}+u\sin\theta\hat{j}$

$\sf:\implies\:\vec{u}=40(\cos 30^{\circ})\hat{i}+40(\sin 30^{\circ})\hat{j}$

$\sf:\implies\:\vec{u}=40\left(\dfrac{\sqrt{3}}{2}\hat{i}+\dfrac{1}{2}\hat{j}\right)$

$\bf:\implies\:\vec{u}=20\sqrt3\hat{i}+20\hat{j}$

❖ Final velocity of projectile :

$\sf:\implies\:\vec{v}=\vec{u}+\vec{a}t$

$\sf:\implies\:\vec{v}=(20\sqrt3\hat{i}+20\hat{j})-(10\hat{j}\times 3)$

$\bf:\implies\:\vec{v}=20\sqrt{3}\hat{i}-10\hat{j}$

❖ Change in velocity :

$\sf:\implies\:\Delta\vec{v}=\vec{v}-\vec{u}$

$\sf:\implies\:\Delta\vec{v}=-30\hat{j}$

Magnitude = 30m/s

∴ (B) is the correct answer!

Answer 2

Given :

Initial speed of projectile = 40 m/s

Angle of projection = 30°

To Find :

Magnitude of change in velocity of projectile till t = 3 after launching.

Solution :

❖ Initial speed of projectile :

$\sf:\implies\:\vec{u}=u\cos\theta\hat{i}+u\sin\theta\hat{j}$

$\sf:\implies\:\vec{u}=40(\cos 30^{\circ})\hat{i}+40(\sin 30^{\circ})\hat{j}$

$\sf:\implies\:\vec{u}=40\left(\dfrac{\sqrt{3}}{2}\hat{i}+\dfrac{1}{2}\hat{j}\right)$

$\bf:\implies\:\vec{u}=20\sqrt3\hat{i}+20\hat{j}$

❖ Final velocity of projectile :

$\sf:\implies\:\vec{v}=\vec{u}+\vec{a}t$

$\sf:\implies\:\vec{v}=(20\sqrt3\hat{i}+20\hat{j})-(10\hat{j}\times 3)$

$\bf:\implies\:\vec{v}=20\sqrt{3}\hat{i}-10\hat{j}$

❖ Change in velocity :

$\sf:\implies\:\Delta\vec{v}=\vec{v}-\vec{u}$

$\sf:\implies\:\Delta\vec{v}=-30\hat{j}$

Magnitude = 30m/s

∴ (B) is the correct answer!