Question

A object is projected from ground with velocity u and angle $\theta$.

After what time will the final velocity become perpendicular to the initial velocity ?

#Revision_Q19


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Answer 1

☆Concept:

》As the object is projected with some velocity in the presence of constant acting downward acceleration(gravity), it will result a projectile motion.

》After time interval 't', the vectors of the final and initial velocities considered will be in such a way,

that on extension they appear to be perpendicular meeting from any particular random axis.

☆Solution:

see the solution with refer to above figure.

• initial velocity = u = $\rm u_{x} \hat{i} + u_{y} \hat{j}$

• final velocity = v = $\rm v_{x} \hat{i} + v_{y} \hat{j}$

》Let's first find the final velocity after time 't' of projection:

• $\rm v_{x} = u_{x}$ ..(no acceleration in x-dirn.)

• $\rm v_{y} = u_{y} - gt$ ...(equation of motion)

》For condition to be perpendicular, their dot product should be zero(where the angle is in the function of cos, and cos90°=0)

• $\rm \vec{v}. \vec{u} = 0$

• $\rm (v_{x} \hat{i} + v_{y} \hat{j}).( u_{x} \hat{i} + u_{y} \hat{j})=0$

• $\rm v_{x}×u_{x} + v_{y}×u_{y}=0$

• $\rm u_{x}² + (u_{y} -gt)(u_{y})=0$

• $\rm u_{x}² + u_{y}² -u_{y}gt=0$

• $\rm u²= u_{y}gt$

• $\rm u² = usin \theta gt$

• $\rm u = sin \theta gt$

》$\boxed{\bf{ t= \dfrac{u}{gsin \theta }}}$

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▪︎note:

• one thing we can mark here that, during the course of motion it started with angle $\theta$ and ended with $\theta$ making on ground.
• so, whatever the time t time given, the angle of perpendicular velocity $\alpha$, will never be greater than initial $\theta$ taken.

Answer 2

Answer:

$\rm t = \dfrac{u}{g sin \theta}$

Explanation:

$\sf Initial \ velocity: \ \rm \overrightarrow{u} = ucos \theta \hat{i} + usin \theta \hat{j} \\ \\ \sf Final \ velocity: \ \rm \overrightarrow{v} = \overrightarrow{u} + \overrightarrow{a} t \\ \\ \sf Acceleration \ due \ to \ gravity: \ \rm \overrightarrow{a} = - g \hat{j} \\ \\ \rm \implies \overrightarrow{v} = (ucos \theta \hat{i} + usin \theta \hat{j} ) + ( - g \hat{j})t \\ \rm = ucos \theta \hat{i} + (usin \theta - gt) \hat{j}$

If final velocity is perpendicular to initial velocity;

$\rm \overrightarrow{v}. \overrightarrow{u} = 0$

$\therefore$

$\rm \implies [ucos \theta \hat{i} + (usin \theta - gt)\hat{j} ].[ucos \theta \hat{i} + usin \theta \hat{j}] = 0 \\ \\ \rm \implies {u}^{2} {cos}^{2} \theta + {u}^{2} {sin}^{2} \theta - ugtsin \theta = 0 \\ \\ \rm \implies {u}^{2} ( {sin}^{2} \theta + {cos}^{2} \theta) - ugtsin \theta = 0 \\ \\ \rm \implies {u}^{2} - ugtsin \theta = 0 \\ \\ \rm \implies u(u - gtsin \theta) = 0 \\ \\ \rm \implies u - gtsin \theta = 0 \\ \\ \rm \implies gtsin \theta = u \\ \\ \rm \implies t = \dfrac{u}{g sin \theta}$