Question

A body is thrown with a speed of 20m/s at an angle of 60 with the horizontal find the time gap between the two positions

Answer 1

Explanation:

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Answer 2

$\frac{3}{2}$1.15 sec is the time gap between the two positions of body.

Explanation:

As angle made with horizontal velocity remains equal

∴ Vi cos 60° = Vr cos 30°

  20 × 1/2 = Vr × √3/2

    20/√3 = Vr

Hence, V = Vr sin 30°

= 20/√3 × 1/2

= 10/√3 m/s

time gap = time when vertical upward velocity

V = u - gt

-10/√3 = 10/√3 - 10t

2/√3 = t

t = 1.15 sec.

Your question is incomplete but most probably your full question was A body is thrown with a speed of 20m/s at an angle of 60 with the horizontal find the time gap between the two positions of body where velocity of body makes an angle of 30° with horizontal.