Question

a projectile is thrown at an angle of 30 degree with a velocity of 10 metre per second the change in velocity during the time interval in which it reaches the highest point is
(1)10m/s (2) 5m/s (3)5✓3m/s. (4)10✓3m/s

Answer 1

Answer:

Option 2 will be the answer

Answer 2

Given:

• Velocity of projection = 10 m/s
• Angle of projection = 30°.

To find:

• Velocity difference in the time interval till it reaches max height ?

Calculation:

We know that :

• In a projectile motion, the velocity component can be divided into two perpendicular velocity vectors along X and Y axis.

Now, at the time of projection :

$\vec{u} = 10 \cos( {30}^{ \circ} ) \hat{i} + 10 \sin( {30}^{ \circ} ) \hat{j}$

After reaching max height, the projectile only has the horizontal component (X axis) of velocity :

$\vec{v} = 10 \cos( {30}^{ \circ} ) \hat{i}$

Now, velocity difference is :

$\vec{u} - \vec{v} = 10 \sin( {30}^{ \circ} ) \: \hat{j}$

$\implies | \vec{u} - \vec{v}| = 10 \times 0.5$

$\implies | \vec{u} - \vec{v}| = 5 \: m {s}^{ - 1}$

So, velocity difference is 5 m/s (option 2)