Question

For a projected body the minimum and maximum velocities are 15 m/s and 30m/s then the angle of projection is
please answer it fast i am going to write exam​

Answer 1

Answer:

The minimum velocity occurs when Vy = 0 since Vx is constant

V = (Vx^2 + Vy^2)^1/2  

Vx = 15 (at the top where Vy = 0)

V^2 = 900 = Vx^2 + Vy^2 = 225 + Vy^2

Vy = (900 - 225)^1/2 = 26 m/s  at launch where Vy is maximum

Vy / Vx = 26 / 15 = tan theta

theta = 60 deg

Answer 2

Answer:

60 degrees is the answer for this one