What is the angle of projection for which the range and maximum height become equal?
Correct answer: B. tan-1(4)
- A. tan-1(1/4)
- B. tan-1(4)
- C. cos-1(1/4)
- D. sin-1(1/4)
Explanation
Let's denote the angle of projection as \( \theta \) and use the formulas for the range (R) and maximum height (H) in projectile motion.The formulas are:Range: R = (v2 sin2θ) / gMaximum Height: H = (v2 sin2θ) / 2gFor the range and maximum height to be equal, we set R = H:(v2 sin2θ) / g = (v2 sin2θ) / 2gBy canceling out common terms:2 sin(2θ) = sin2(θ)Solving for θ:2 sin(2θ) = sin2(θ)2 (2 sin(θ) . cos(θ)) = sin2(θ)4 sin(θ) . cos(θ) = sin2(θ)4 cos(θ) = sin(θ)tan(θ) = 4 Therefore, the angle of projection for which the range and maximum height become equal is θ = tan-1(4).
Report an error
The more specific you are, the faster it gets fixed. A source beats an opinion.
Prefer email? [email protected]
About Force and Motion
Displacement, velocity and acceleration describe motion, including the horizontal and vertical components of projectile motion. The chapter connects these quantities with Newton's laws, momentum and impulse, then applies conservation of momentum to collisions, distinguishing elastic collisions from inelastic ones.
Practise Force and Motion
1,416 free Force and Motion MCQs from Physics, each with the correct answer and an explanation. Unlimited attempts, no account needed.
Discussion
Stuck on an option, or know a faster way to get there? Ask or explain it here.
No comments yet. Be the first to explain this one.
Exams that ask Physics questions like this
Physics is on 16 papers prepared for on TestUstad, and all of them draw the same bank, so this question is worth knowing for every one of them.
More Force and Motion questions
Which of the following is not an-example of projectile motion:
Taking off by rocket can be explained by:
If an iron ball and a wooden ball of the same radius are released from the height 'h' in a vacuum, then the time taken by both of them to reach the ground will be:
When two bodies move toward each other at constant speeds, the distance between them decreases at the rate of 6m/sec. If they move in the same. direction, the distance between them increases at the rate of 4M/sec. Then their speeds are:
Distance covered by a freely falling body in 2 seconds will be:
The distance covered by a body with uniform acceleration in time t starting from rest is