Question

For a projectile ( range)² is 48 times of (maximum height)² obtained. The angle of projection is:

a) 60°

b) 30°

c) 45°

d) 75°

Answer 1

Given:-

(R)^2= 48.(h)^2…………..(1)

Let angle of projection = a° with horizontal.

Initial velocity = u

vertical component of velocity = u.sina.

Horizontal component of velocity= u cosa.

Max.height =h

Range =R

Vertical motion:-

v^2 =u^2 -2gh

(0)^2 =(u.sina)^2 -2gh

h= u^2.sin^2a/2g……………..(2)

v=u-g.t

0 = usina-gt

t= u.sin a/g

Total time of flight (T) =2.t. =2u.sin a/g……..(3)

.Horizontal motion:-

Range(R)=horizontal component of velocity×T

R = ucos a ×2u.sin a/g

R = u^2.(2sin a.cos a)/g……………….(4)

Given:-

R^2 = 48.h^2

u^4.(4.sin^2 a.cos^2a)/g^2 =48.u^4.sin^4a/4g^2

cos^2a =3.sin^2a.

1/3= sin^2a/cos^2a

tan^2a = 1/3.

tan a = 1/3^1/2 =tan 30°

or a=30° , Answer

so the angle of projection is 30 degree