AP EAMCET Engineering 21 Apr 2019 Shift 1 Paper

The expression for the maximum height of the projectile is given by,
H=‌

u2sin2θ

2g

..............(1)
The expression for the maximum range of the projectile is given by
R=‌

u2‌sin‌2‌θ

g

The ratio of the maximum height to the range of the projectile is calculated as,
‌

H

R

=‌

‌ u2sin2θ 2g

u2‌sin‌2‌θ g

=‌

sin3θ

2‌sin‌2‌θ

Substitute the given values in the above equation.
‌

H

R

=‌

sin3θ

2‌sin‌2‌θ

‌

3

4

=‌

sin3θ

2‌sin‌2‌θ

3‌sin‌2‌θ=2sin2θ
tan‌θ=3
From the geometry of the triangle,
sin‌θ=‌

3

√10

.............(2)
From the equation (1)
‌

u2sin2θ

2g

=3
u2sin2θ=3×2g
Substitute the values in the above equation.
u2(‌

3

√10

)2=3×2(10)
u2‌‌=‌

100×2

3

u‌‌=10√‌

2

3

ms−1