JEE Main 1-Sep-2021 Shift 2 Solved Paper

Here, two projectiles are projected at angles 42° and 48° withsame initial velocity.
As we know the expression of range of projectile,
Range =

u2‌sin‌2‌θ

g

At θ1 = 42°
Range, R1=

u2‌sin‌2(42)°

g

=

0.99u2

g

At θ2 = 48°
Range, R2=

u2‌sin‌2(48)°

g

=

0.99u2

g

The range of the projectile is same for the two projectiles.
Therefore, R1=R2
Now, as we know the expression of height of the projectile,
Hmax=

u2‌sin‌θ

2g

At 42°,
Hmax=H1=

u2‌sin‌42°

2g

=

0.669u2

2g

At 48°,
Hmax=H2=

u2‌sin‌48°

2g

=

0.743u2

2g

Higher the value of q higher the value of maximum height.Therefore, H1<H2.