Projection Under uniform Acceleration

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Question

A projectile is thrown at an angle of 30° with a velocity of 10 m/s. the change in velocity during the time interval in which it reaches the highest point is

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Solution

Initial velocity of projection $\large \overrightarrow u = \left( {10\cos {{30}^0}\widehat i + 10\sin {{30}^0}\widehat j} \right)m/s$
$\large = \left( {5\sqrt 3 \widehat i + 5\widehat j} \right)m/s$
Horizontal component of velocity remains unchanged. At the highest point velocity is horizontal and $\large \overrightarrow V = 5\sqrt 3 \widehat {i\;}m/s$
$\large \therefore$ Magnitude of change in velocity $\large = \left| {\overrightarrow {\Delta V} } \right| = \left| {\overrightarrow v - \overrightarrow u } \right| = 5m/s$

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Projection Under uniform Acceleration

Remember concepts with our Masterclasses.

A projectile is thrown at an angle of 30° with a velocity of 10 m/s. the change in velocity during the time interval in which it reaches the highest point is

Initial velocity of projection Horizontal component of velocity remains unchanged. At the highest point velocity is horizontal and Magnitude of change in velocity

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