Q. A ball is thrown upward with velocity $v$ and at an angle $\theta $ with the ground. What will be velocity of the ball after $t$ seconds? (Given mass of the ball $=m$ )
NTA AbhyasNTA Abhyas 2020
Solution:
As we know that for a projectile motion the instantaneous velocity after $tsec$ is given by:
$\left|\overset{ \rightarrow }{V_{t}}\right|=\sqrt{v_{x}^{2} + v_{y}^{2}}$
Where $v_{x}=vcos \theta $ is the horizontal component of velocity.
$v_{y}=vsin \theta -gt$ is the vertical component of velocity .
Hence putting these values in the above equation,we get:
$\Rightarrow \left|v_{t}\right|=\sqrt{\left(v \, c o s \theta \right)^{2} + \left(v sin \theta - g t\right)^{2}}$
$\Rightarrow \left|v_{t}\right|=\sqrt{v^{2} + g^{2} t^{2} - 2 v sin \theta \, g t \, }$
$\left|\overset{ \rightarrow }{V_{t}}\right|=\sqrt{v_{x}^{2} + v_{y}^{2}}$
Where $v_{x}=vcos \theta $ is the horizontal component of velocity.
$v_{y}=vsin \theta -gt$ is the vertical component of velocity .
Hence putting these values in the above equation,we get:
$\Rightarrow \left|v_{t}\right|=\sqrt{\left(v \, c o s \theta \right)^{2} + \left(v sin \theta - g t\right)^{2}}$
$\Rightarrow \left|v_{t}\right|=\sqrt{v^{2} + g^{2} t^{2} - 2 v sin \theta \, g t \, }$