Question

When a body is projected at an angle of theta the maximum vertical height attained is onefourth?

Answer 1

Explanation:

The angle of projection for which the maximum height of the projectile is equal to the horizontal range has to be determined.

Answer 2

H = 8u'2 / 17g

Given:

 A body is projected at an angle of theta.

To Find:

The maximum vertical height attained is onefourth?

Solution:

Let's consider u = initial speed of the projectile

8 = angle of projection According to the question,

H = R

u'sin'∅/2u u'sin2∅/2u  

sin∅.sin∅ = 2 x 2sin∅.cos∅

sin∅ = 4cos∅

tan∅ = 4

As we know that,

1 + tan'2 ∅ = sec'2  ∅

1/cos;2  ∅ =   l + 16 = 17

cos'2  ∅  = 1/17

sin'2 ∅ = l —cos'2  ∅

sin'2  ∅ = l-1/17    = l6/17

maximum height H

H  =  u/2g  .  16/ 17

H = 8u'2 / l7u

H = 8u'2 / 17g

#SPJ3