Question

a body is projected with initial velocity of 10root 3m/sec making an angle 30 with the horizontal the velocity of the particle at the highest point of the trajectory is (a)15 m/sec B ​

Answer 1

Answer:

15 m/s                  [∵ v = u cosθ]

Explanation:

Given:

u = 10$\sqrt{3}$ m/s

θ = 30°

At maximum height vertical component of velocity is zero

⇒ only horizontal component is present

⇒ v = u cosθ

     = 10$\sqrt{3}$ × cos30°

     = 10$\sqrt{3}$ × $\frac{\sqrt{3} }{2}$

     = 5 × 3 m/s

     = 15 m/s

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