A projectile is projected with velocity of 25 m s - 1 at an angle θ with the horizontal. After t seconds its inclination with horizontal becomes zero. If R represents horizontal range of the projectil

Question

A projectile is projected with velocity of 25 m s - 1 at an angle θ with the horizontal. After t seconds its inclination with horizontal becomes zero. If R represents horizontal range of the projectile, the value of θ will be : [use use g = 10 m s - 2 ]

TestHub · Accepted Answer

From the information given in the question, after time $t$ the inclination of the projectile becomes zero i.e. the projectile is at the highest point. Therefore,

$t = \frac{T}{2} = \frac{u \sin θ}{g}$ .

The rang of the projectile $R = \frac{2 u^{2} \sin θ \cos θ}{g}$ .

Now,

$\frac{R}{t^{2}} = \frac{2 u^{2} \sin θ \cos θ}{g} \times \frac{g^{2}}{u^{2} \sin^{2} θ} = 2 g \cot θ \Rightarrow θ = \cot^{- 1} (\frac{R}{20 t^{2}})$

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