TS EAMCET 5-Aug-2021 Shift 2 Question Paper

Given,
Let speed of projection be u1,
Speed of projectile at maximum height be u2
R is range, H is maximum height,
and u1=2u2
As we know that,
H=‌

u12sin2θ

2g

and ‌‌R=‌

u12‌sin‌2‌θ

g

where, θ= angle of projection,
and g= acceleration due to gravity.
From given diagram,u2=u1‌cos‌θ
⇒‌‌u2=2u2‌cos‌θ⇒cos‌θ=‌

1

2

⇒‌‌cos‌θ=cos‌60∘⇒θ=60∘
On dividing Eq. (ii) by Eq. (i), we get
‌

R

H

=‌

u12‌sin‌2‌θ

g

×‌

2g

u12sin2θ

=‌

2‌sin‌θ‌cos‌θ

sin2θ

⋅2=‌

2‌cos‌θ

sin‌θ

⋅2=4‌cot‌θ
As, ‌‌θ=60∘
∴‌‌‌

R

H

=4‌cot‌60∘=4×‌

1

√3

=‌

4

√3