AP EAMCET · PHYSICS · Motion In Two Dimensions

A body is projected at an angle of $60 °$ with the horizontal such that the vertical component of its initial velocity is $40 m s^{- 1}$ . The magnitude of velocity of the projectile at one quarter of its time of flight is nearly (Acceleration due to gravity $= 10 m s^{- 2}$ )

A $3.54 m s^{- 1}$
B $35.40 m s^{- 1}$
C $30.54 m s^{- 1}$
D $34.5 m s^{- 1}$

Verified Solution

Answer & Solution

Correct Answer

(C) $30.54 m s^{- 1}$

Step-by-step Solution

Detailed explanation

Suppose initial velocity is v and angle of projectionθ is given 60°. Given, Vertical component of initial velocity v sin θ=40 m s-1 ⇒ v=46.19 m s-1 Time of flight T=2 v sin θg=2×4010=8 s…

Apply the relevant identity from the chapter and substitute the values established in the previous step, keeping the original constraint in view.

Use the simplified expression to isolate the unknown, then plug the result back into the question setup to confirm the working is internally consistent.

Cross-check against a boundary or limiting case. Both the dimensional balance and the qualitative behaviour at the extreme should agree with the candidate answer.

Combine the steps above to identify the correct option. The remaining choices each fail at least one of the consistency, dimensional, or boundary checks.

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A body is projected at an angle of 60° with the horizontal such that the vertical component of its initial velocity is 40 m s-1. The magnitude of velocity of the projectile at one quarter of its time of flight is nearly (Acceleration due to gravity =10 m s-2)

Step-by-step Solution

A body is projected at an angle of $60 °$ with the horizontal such that the vertical component of its initial velocity is $40 m s^{- 1}$ . The magnitude of velocity of the projectile at one quarter of its time of flight is nearly (Acceleration due to gravity $= 10 m s^{- 2}$ )