JEE PHYSICS MCQ: MOTION IN PLANE

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A hiker stands on the edge of a cliff

490 \mathrm{~m}

above the ground and throws a stone horizontally with an intial speed of

15 \mathrm{~m} / \mathrm{s}

. Neglecting air resistance. Find the time taken by the stone to reach the ground, and the speed with which it hits the ground

\left({g}=9.8 \mathrm{~m} / \mathrm{s}^2\right)

View answer and explanation

Correct option: B

10 \mathrm{~sec}, 99 \mathrm{~m} / \mathrm{s}

Option B (

10 \mathrm{~sec}, 99 \mathrm{~m} / \mathrm{s}

) is correct because it accurately calculates the time of flight and final velocity of the stone using kinematic equations.

More Questions from MOTION IN PLANE

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

\mathrm{\vec A}\,and\,\mathrm{\vec B}

0^{\circ}

90^{\circ}

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

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

\mathbf{a}

\mathbf{b}

\theta

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

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

50 \mathrm{~m}

{g}=9.8 \mathrm{~ms}^{-2}