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For a angular projection given to a projectile, find : (i) Maximum height, (ii) Time of flight, (iii) Horizontal range.

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(i)Maximum height – It is the maximum vertical height attained by the object above the point of projection during its flight denoted by ℎ.

y= sin θ, ay= −g, y0 = 0

y= ℎ, t= T/2 =\({usinθ}/{g}\)

Using relation,

y= y0+uyt+\(\frac{1}{2}u_yt^2\),

We have ℎ = 0 +usin θ × \(\frac{ sin θ}{g}\) + \(\frac{1}{2}\) (−g) (\(\frac{ sin θ}{g}\) ) 2

or ℎ = \(\frac{u^2}{g}\) sin2θ − \(\frac{1}{2}\)\(u^2\frac{ sin^2θ}{g}\) 

ℎ = \(\frac{u^2sin^2θ}{2g}\)

 (ii)Time of flight – The total time for which the projectile is in flight, taking vertical downward motion of object from O to C.

u= vsin θ, a = −9

Let time taken to complete the trajectory = T

As the projectile is reaching the same level of projection vertical displacement, = 0

We have, s=ut+ \(\frac{1}{2}at^2\)

0 = usin θ.T – \(\frac{1}{2}\)gT2

T = \(\frac{2usin θ}{g}\) 

(iii) Horizontal range – It is the horizontal distance travelled by projectile during its flight i.e. horizontal distance covered by object while going from O to C. Distance = velocity × time

= µcos θ × T

= \(\frac{ \mu cosθ ×2u sin θ}{g}\)

= \(\frac{u^2(2 sin θ cos θ)}{g}\)

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