JEE PHYSICS MCQ: MOTION IN PLANE

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A cricket ball thrown across a field is at heights

h_1

and

h_2

from the point of projection at times

t_1

and

t_2

respectively after the throw. The ball is caught by a fielder at the same height as that of projection. The time of flight of the ball in this journey is

View answer and explanation

Correct option: C

\left(\frac{h_1 t_2^2-h_2 t_1^2}{h_1 t_2-h_2 t_1}\right)

Option C (

\left(\frac{h_1 t_2^2-h_2 t_1^2}{h_1 t_2-h_2 t_1}\right)

) is correct because it's derived from using kinematic equations and solving for the time of flight based on the given heights and times.

More Questions from MOTION IN PLANE

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Given,

\mathrm{\vec P} = \mathrm{\vec A} + \mathrm{\vec B}\,\,\mathrm{and}\,P = A + B.

The angle between

\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^{\circ}

and

90^{\circ}

between them.
Reason: For two-dimensional or three-dimensional motion, velocity and acceleration of an object works in opposite direction.

\mathrm{If}\,(\mathbf{a}+3 \mathbf{b}) \cdot(7 \mathbf{a}-5 \mathbf{b})=0

and

(\mathbf{a}-4 \mathbf{b}) \cdot(7 \mathbf{a}-2 \mathbf{b})=0

. Then, the angle between

\mathbf{a}

and

\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

\theta

and another ball is thrown at an angle

\left(90^{\circ}-\theta\right)

with the horizontal direction from the same point with velocity

39.2 \mathrm{~ms}^{-1}

. The second ball reaches

50 \mathrm{~m}

higher than the first ball. Find their individual heights. Take

{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.