Question

the potential energy of a projectile at maximum height is 3/4 times kinetic energy of projection .its angle of projection is

Answer 1

Answer: $\bold{60^{\circ}}$


Let us suppose that an object of mass m is projected at an angle $\theta$ with a velocity u.


The Maximum Height of the projectile is given by:

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

So, Potential Energy at Maximum Height would be:

$U = mgH \\ \\ \\ \implies U = m \times \cancel{g} \times \frac{u^2\sin^2\theta}{2\cancel{g}} \\ \\ \\ \implies U = \frac{mu^2\sin^2\theta}{2}$


Also, since the object was projected with a velocity u, the Kinetic Energy of Projection would be:

$K = \frac{1}{2}mu^2$


Now, we are given that the Potential Energy of the projectile at maximum height is 3/4 times the Kinetic Energy of projection. So we can write:

$U = \frac{3}{4}K \\ \\ \\ \implies \frac{\cancel{m} \, \cancel{u^2}\sin^2\theta}{\cancel{2}} = \frac{3}{4} \times \frac{1}{\cancel{2}}\cancel{m} \, \cancel{u^2} \\ \\ \\ \implies \sin^2\theta = \frac{3}{4} \\ \\ \\ \implies \sin \theta = \frac{\sqrt{3}}{2} \\ \\ \\ \implies \boxed{\theta = 60^{\circ}}$


Thus, The Angle of Projection is $\bold{60^{\circ}}$

Answer 2

k.e= ½m(vcosx)²
k.e=½mv²cos²x
Also,
v²=u²+2as
0=v²sin²x +2(-g)h
h=(v²sin²x)/2g
so potential energy
=mgh
=mg×{(v²sin²x)/2g}
=½(mv²sin²x)
now a/q,
p.e=¾(k.e)
k.e= 4/3(p.e)
k.e=4/3[½(mv²sin²x)]
k.e= 2/3(mv²sin²x)
½mv²cos²x=⅔(mv²sin²x)
(sin²x)/(cos²x)=3/4
tan²x=3/4
tanx=√3/2
x=tan inverse(√3/2)