Q21.A body is projected from the ground at an angle of 45° with the horizontal. Its velocity after 2 s is 20 m s−1 . The maximum height reached by the body during its motion is _____ m . (use g = 10 m s−2 )
What This Question Tests
This question tests the application of kinematic equations for projectile motion to find the initial velocity components and then calculate the maximum height reached.
Concepts Tested
Formulas Used
vx = ux
vy = uy - gt
v = sqrt(vx^2 + vy^2)
H = uy^2 / (2g)
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📋 Question Details
- Chapter
- Kinematics
- Topic
- Projectile Motion
- Year
- 2022
- Shift
- 24 Jun Shift 2
- Q Number
- Q21
- Type
- Numerical
- NCERT Ref
- Class 11 Physics Ch 4: Motion in a Plane
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