The velocity of projection of an oblique projectile is 𝐯⃗ = 3𝐒̂+ 2𝐣̂(in msβˆ’1 ). The speed of the projectile at the highest point of the trajectory is

Question : The velocity of projection of an oblique projectile is 𝐯⃗ = 3𝐒̂+ 2𝐣̂(in msβˆ’1 ). The speed of the projectile at the highest point of the trajectory is

(A) 3ms^{-1}

(B) 2ms^{-1}

(C) 1ms^{-1}

(D) zero

Answer : option (A)

Solution :

The velocity of projection of an oblique projectile is 𝐯⃗ = 3𝐒̂+ 2𝐣̂(in msβˆ’1 ). The speed of the projectile at the highest point of the trajectory is
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