JEE Main 27-Aug-2021 Shift 2 Solved Paper

Given, initial speed, u=25ms−1
Angle of projection, θ=45∘ Let, the maximum height =H
Time taken to reach, H=t
Acceleration due to gravity g=10ms−2
Velocity at maximum height along Y-axis,
vy=0ms−1
As we know that, for motion along
Y-axis,
∵‌vy2−uy2=2gH
∴‌02−(25‌sin‌45∘)2=−2×10×H
⇒−312.5=−20H
∵‌H=15.625m
∴‌H=‌

1

2

gt2
∴‌=√‌

2H

g

=√‌

2×15.625

10

‌=1.767=1.77s