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3-2.Motion in Plane
easy
કણ ને $u$ વેગથી સમક્ષિતીજ સાથે ખૂણે $\theta$ ફેકવામા આવે તો મહત્તમ ઊચાઇએ તેના વેગમા કેટલો ફેરફાર થાય?
A
$u\, cos\theta$
B
$u$
C
$u \,sin\theta$
D
$(u \,cos\theta \, -\,u)$
Solution
At the time of projection, the particle's speed is $u$ and the particle's speed at the
$\Delta u = u – u \cos \theta$
Now, the initial velocity is,
$\overrightarrow{ u }= u \cos \theta \hat{ i }+ u \sin \theta \hat{ j }$
and the final velocity (at highest point) is,
$\overrightarrow{ v }= u \cos \theta \hat{i}$
Therefore, the change in velocity is,
$\Delta \overrightarrow{ v }=\overrightarrow{ u }-\overrightarrow{ v }=u \cos \theta \hat{ i }+u \sin \theta \hat{ j }- u \cos \theta \hat{ i }=u \sin \theta \hat{ j }$
$\Rightarrow|\Delta \overrightarrow{ v }|=\operatorname{usin} \theta$
Standard 11
Physics