TS EAMCET 10-Sep-2020 Shift 2 Solved Paper

The given situation is shown in the following figure
Angle of projection, θ=45∘ The elevation angle of the projectile at its highest point as seen from the point of projection is α.
∴ In △POM
tan‌α=‌

H

R∕2

=‌

‌ u2sin2θ 2g

u2‌sin‌2‌θ 2g

=‌

sin2θ

sin‌2‌θ

=‌

sin245∘

sin‌90∘

=‌

( 1 √2 )2

1

⇒tan‌α=‌

1

2

⇒‌‌α=tan−1(‌

1

2

)