NEET UG PHYSICS MCQ: MOTION IN PLANE

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Application +4 | -1

Two towers

A

and

B

each of height

20 \mathrm{~m}

are separated by a distance

200 \mathrm{~m}

. A body thrown horizontally from the tower

A

with a velocity

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

towards the tower

B

hits the ground at a point

P

. Similar body projected horizontally from the tower

B

towards

A

with same velocity hits the ground at

Q.

The uniform velocity with which a truck can move from

P

Q

in 4 seconds is

\left(g=10 \mathrm{~m} / \mathrm{s}^2\right)

View answer and explanation

Correct option: D

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

Option D is correct because the horizontal distance covered depends on the time of flight, which is determined by the height of the tower.

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\mathrm{\vec a}\,and\,\mathrm{\vec b}

\mathrm{\vec a}\,and\,\mathrm{\vec b}

\overrightarrow P \,\,.\,\,\overrightarrow Q = 0\,and\,\overrightarrow P \,\, \times \,\,\overrightarrow Q = 0

\overrightarrow Q \,and\,\overrightarrow R

\vec{A}=3 \hat{i}+4 \hat{j}

\vec{B}=6 \hat{i}+8 \hat{j}

A

B

\overrightarrow{\mathrm{A}}

\overrightarrow{\mathrm{B}}

{\cot ^{ - 1}}\left( 2 \right)