The time of flight of a projectile is maximum when angle of projection is MCQ with Answer and Explanation

The time of flight of a projectile is maximum when angle of projection is
A. 90°
B. 45°
C. 30°
D. 60°
Answer: Option A
Solution (By JKSSB Mock Tests)
T = 2u sinθ/g, max when sinθ=1 => θ=90° (vertical).

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Practice More Physics Questions

Question #1
What is the excess pressure strictly inside a soap bubble of radius R and surface tension T?
A. T / 2R
B. 4T / R
C. 2T / R
D. T / R

Correct Answer: Option B


Explanation:
A liquid drop has only one free surface, so its excess pressure is P = 2T/R. However, a soap bubble in the air has two completely free surfaces (an inner one and an outer one). Because both surfaces exert surface tension forces inward, the total excess pressure inside a soap bubble is exactly double: P = 4T/R.

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Question #2
A body is dropped from a height of 20 m. Its velocity on reaching ground (g=10) is
A. 10 m/s
B. 40 m/s
C. 20 m/s
D. 14.14 m/s

Correct Answer: Option C


Explanation:
v² = 2gh = 2×10×20=400, v=20 m/s.

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Question #3
The area under acceleration-time graph represents:
A. Change in velocity
B. Velocity
C. Distance
D. Displacement

Correct Answer: Option A


Explanation:
Acceleration a = dv/dt ⇒ dv = a·dt. Integrating: ∫dv = ∫a·dt ⇒ Δv = area under a-t graph. Slope of v-t gives acceleration; area under a-t gives velocity change. Memory aid: 'a-t graph area = Δv; v-t graph area = displacement'. Graph interpretation skill essential for motion analysis in competitive exams.

This question belongs to: Science Physics