Question

Class 11th physics vector related motion in a plane

Answer 1

MAXIMUM VERTICAL HEIGHT:-

The velocity of the particle is horizontal, The vertical component of the particle is Horizontal. the vertical component of velocity is thus , zero at the highest point.the maximum height is the y - coordinate of the particel when the vertical component of velocity is zero

Derive the value of maximum height

We have ,

$\rm \to \: v_y = u_y - gt$

taking component we get

$\to \rm \: u_y = u \sin\theta$

$\rm \: \to \: v_y = 0$

we get

$\rm \to \: v_y = u \sin\theta - gt$

At maximum height

$\rm \to \: 0 = u \sin\theta - gt$

or

$\boxed{ \rm \: t = \dfrac{u \sin \theta }{g} }$

Now the maximum height is

$\rm \: \to \: H = u_yt - \frac{1}{2} gt {}^{2}$

$\rm = \: (u \sin\theta) \bigg( \dfrac{u \sin \theta}{g} \bigg) - \dfrac{1}{2} g \bigg( \dfrac{u \sin \theta }{g} \bigg) {}^{2}$

$\rm = \dfrac{ {u}^{2} \sin {}^{2} \theta}{g} - \dfrac{1}{2} \dfrac{ {u}^{2} \sin {}^{2} \theta }{g}$

$\rm = \dfrac{ {u}^{2} \sin {}^{2} ( \theta) }{2g}$

$\rm \: maximum \: height \: (H) =\dfrac{ {u}^{2} \sin {}^{2} ( \theta) }{2g}$

Answer 2

Explanation:

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Answer♡

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