Question

A projectile is projected from ground with velocity of 20 m/s at angle of 53° with horizontal. At t = t0, its velocity becomes perpendicular to its initial velocity. If v be the speed of particle at t = t0, then [g = 10 m/s2]

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Answer 1

Explanation:

$\huge\bold{\red{2.50 s}}$

Answer 2

Given:

A projectile is projected from ground with velocity of 20 m/s at angle of 53° with horizontal. At t = t0, its velocity becomes perpendicular to its initial velocity.

To find:

Correct option ?

Calculation:

Initial velocity vector :

$\vec{u} = u \cos( \theta) \hat{i} + u \sin( \theta) \hat{j}$

Final Velocity vector after time t :

$\vec{v} = u \cos( \theta) \hat{i} + \{u \sin( \theta) - gt \} \hat{j}$

Now , since these two vectors are perpendicular, we can say :

$\therefore \: \vec{u} \: . \: \vec{v} = 0$

$\implies {u}^{2} { \cos}^{2} ( \theta) + {u}^{2} { \sin}^{2} ( \theta) - ugt \sin( \theta) = 0$

$\implies {u}^{2} \bigg \{{ \cos}^{2} ( \theta) + { \sin}^{2} ( \theta) \bigg \} - ugt \sin( \theta) = 0$

$\implies {u}^{2} - ugt \sin( \theta) = 0$

$\implies u - gt \sin( \theta) = 0$

$\implies u = gt \sin( \theta)$

$\implies t = \dfrac{u}{g\sin( \theta) }$

$\implies t = \dfrac{20}{g\sin( {53}^{ \circ} ) }$

$\implies t = \dfrac{20}{10\sin( {53}^{ \circ} ) }$

$\implies t = \dfrac{2}{\sin( {53}^{ \circ} ) }$

$\implies t = \dfrac{2}{ (\frac{4}{5} ) }$

$\implies t = 2.5 \: sec$

So, time taken for the vectors wo be perpendicular is 2.5 sec.

$\therefore \: \vec{v} = u \cos( \theta) \hat{i} + \{u \sin( \theta) - gt \} \hat{j}$

$\implies \: \vec{v} = 20\cos( {53}^{ \circ} ) \hat{i} + \{20 \sin( {53}^{ \circ} ) - (10)(2.5) \} \hat{j}$

$\implies \: \vec{v} = 20( \dfrac{3}{5} )\hat{i} + \{20( \dfrac{4}{5} ) - 25 \} \hat{j}$

$\implies \: \vec{v} = 12\hat{i} + \{9 - 25 \} \hat{j}$

$\implies\:\vec{v} = 12\hat{i} -16\hat{j}$

$\implies \: | \vec{v}| = \sqrt{ {12}^{2} + {16}^{2} }$

$\implies \: | \vec{v}| = \sqrt{400 }$

$\implies \: | \vec{v}| =20 \: m {s}^{ - 1}$

So, Velocity at that time will be 20 m/s.