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JEE PHYSICS MCQ: MOTION IN PLANE

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A projectile is thrown upward with a velocity v0v_0 at an angle α\alpha to the horizontal. The change in velocity of the projectile when it strikes the same horizontal plane is
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Correct option: B
2v0sin⁡α2 v_0 \sin \alpha vertically downward
Option B (2v0sin⁡α2 v_0 \sin \alpha vertically downward) is correct because it represents the change in the vertical component of velocity when the projectile returns to the same horizontal plane.

More Questions from MOTION IN PLANE

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Given, P⃗=A⃗+B⃗  and P=A+B.\mathrm{\vec P} = \mathrm{\vec A} + \mathrm{\vec B}\,\,\mathrm{and}\,P = A + B. The angle between A⃗ and B⃗\mathrm{\vec A}\,and\,\mathrm{\vec B} is Assertion: The necessary condition for motion in two dimensions, velocity and acceleration vectors should have any angle between 0∘0^{\circ} and 90∘90^{\circ} between them.
Reason: For two-dimensional or three-dimensional motion, velocity and acceleration of an object works in opposite direction.
If (a+3b)⋅(7a−5b)=0\mathrm{If}\,(\mathbf{a}+3 \mathbf{b}) \cdot(7 \mathbf{a}-5 \mathbf{b})=0 and
(a−4b)⋅(7a−2b)=0(\mathbf{a}-4 \mathbf{b}) \cdot(7 \mathbf{a}-2 \mathbf{b})=0. Then, the angle between a\mathbf{a} and b\mathbf{b} is
At the top of the trajectory of a projectile, the direction of its velocity and acceleration are A ball is thrown at an angle θ\theta and another ball is thrown at an angle (90∘−θ)\left(90^{\circ}-\theta\right) with the horizontal direction from the same point with velocity 39.2 ms−139.2 \mathrm{~ms}^{-1}. The second ball reaches 50 m50 \mathrm{~m} higher than the first ball. Find their individual heights. Take g=9.8 ms−2{g}=9.8 \mathrm{~ms}^{-2}. Assertion: The drover in a vehicle moving with a constant speed on a straight road is in a non-inertial frame of reference.

Reason: A reference frame in which Newton's laws of motion are applicable is non-inertial.
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