A ball is thrown upwards and it returns to ground describing a parabolic path.Which of the following has same value at the time of thrown and time of return?

Kinetic energy of the ball

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Speed of the ball

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Horizontal component of velocity

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Vertical component of velocity

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Solution

The correct options are
A Kinetic energy of the ball
B Speed of the ball
C Horizontal component of velocity
Horizontal component $u_{x}$ of velocity $\to u$ does not change during motion.
Vertical component $v_{y}$ when it returns to ground is $- u_{y}$
$(K E)_{i} = \frac{1}{2} m (u_{x}^{2} + u_{y}^{2})$
$(K E)_{f} = \frac{1}{2} m (u_{x}^{2} + (- u_{y})^{2}) = \frac{1}{2} m (u_{x}^{2} + u_{y}^{2})$
$∴ (K E)_{f} = (K E)_{i}$
Initial speed of the ball: $\sqrt{u_{x}^{2} + u_{y}^{2}}$
Final speed of the ball: $\sqrt{u_{x}^{2} + (- u_{y})^{2}} = \sqrt{u_{x}^{2} + u_{y}^{2}}$
$∴$ speed of the ball is same initially and finally.
vertical component of velocity changes and becomes $- u sin θ$ when it returns to the ground.

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