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3-2.Motion in Plane
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A ball is projected with a velocity, $10 ms ^{-1}$, at an angle of $60^{\circ}$ with the vertical direction. Its speed at the highest point of its trajectory will be$............... ms ^{-1}$
A
$5 \sqrt{3}$
B
$5$
C
$10$
D
Zero
(NEET-2022)
Solution
At highest point only horizontal component of velocity remains $\Rightarrow u _{ x }= u \cos \theta$
$u _{ x }= u \cos \theta =10 \cos 30^{\circ}$
$=5 \sqrt{3} ms ^{-1}$
Standard 11
Physics
Similar Questions
A particle is projected from ground with velocity $u$ at angle $\theta$ from horizontal. Match the following two columns.
| Column $I$ | Column $II$ |
| $(A)$ Average velocity between initial and final points | $(p)$ $u \sin \theta$ |
| $(B)$ Change in velocity between initial and final points | $(q)$ $u \cos \theta$ |
| $(C)$ Change in velocity between initial and final points | $(r)$ Zero |
| $(D)$ Average velocity between initial and highest points | $(s)$ None of the above |
medium
hard
