Question

A ball is projected with initial speed u at an angle θ with the horizontal. the instant when its speed becomes minimum, it is given a sharp impulse due to which it gets additional velocity equal to its 'minimum speed × tanθ' in vertical direction. maximum height reached by it above the ground is

Answer 1

I hope you have understand everything

Answer 2

Answer:

The maximum height above ground is

$\frac{u {}^{2}sin {}^{2}θ }{g}$

Step-by-step explanation:

For a projectile motion,At maximum height of its motion,the components of velocity are

$V _{y} = 0 \: and \: V_{x} = ucosθ$

Therefore,the particle has minimum velocity at maximum height and it is equal to

$V _{ min}=ucosθ$

Now the additional velocity is given as minimum speed x tanθ.Therefore additional velocity is

$ucosθ \times tanθ = usinθ$

Now,the particle is projected with this velocity from its maximum point.Therefore,the maximum height reached by projectile from its new point of projection is

$H' = \frac{ {u {}^{2} sin}^{2}θ }{2g}$

The maximum height reached by ground is sum of initial and new maximum heights.

Initial maximum height is given as

$H = \frac{u {}^{2}sin {}^{2}θ }{2g}$

Therefore Maximum height above ground is

$H _{g} = H' + H = \frac{u {}^{2}sin {}^{2} θ }{2g} + \frac{u {}^{2}sin {}^{2} θ }{2g} \\ = \frac{u {}^{2}sin {}^{2} θ }{g}$

Hence,the maximum height above ground is

$\frac{u {}^{2}sin {}^{2} θ }{g}$

#SPJ3