Question

A projectile time of flight t is related to horizontal range by the (gt)2=2r.The angle of projectile is

Answer 1

(a) For projectile motion,

time of flight

T== 2usinθ / g

On squaring both sides,

T2 = 4u2 sin2θ / g2

or gT2 = 4 u2 sin2 θ/ g--------(i)

and 

horizontal range

R = u2sin2θ /g = 2u2sinθcos θ/g--------(ii)

Dividing (i) by (ii),we get

 gT2/R=2tanθ or gT2=2R.tanθ

(b)

We know,



(c)From the above we have,

 gT2=2R.tanθ and Hmax=(R/4).Tanθ

So  gT2/ Hmax=8

or gT2=8Hmax

Answer 2

Answer:

$\theta=45^{\circ}$    

Explanation:

Let T is the time of flight and R is the horizontal range. The formula for the time of flight is given by :

$T=\dfrac{2v\ sin\theta}{g}$

The formula for the range of a projectile is given by :

$R=\dfrac{v^2\ sin2\theta}{g}$

According to this question,

$gT^2=2R$

$g(\dfrac{2v\ sin\theta}{g})^2=2\times \dfrac{v^2\ sin2\theta}{g}$

On solving the above equation, we find the value of angle of projection is, $\theta=45^{\circ}$

So, the angle of projectile is 45 degrees. Hence, this is the required solution.