Aphysics class conducting a research project on projectile motion constructs a device that can launch a cricket ball. the launching device is designed so that the ball can be launched at ground level with an initial velocity of 28 m/ s at an angle of 30° to the horizontal. calculate the horizontal component of the velocity of the ball: a. initiallyb. after 1.0 sc. after 2.0 s

Explanation:

initial velocity, u = 28 m/s

Angle of projection, θ = 30°

The acceleration in horizontal direction is zero, so the horizontal component of velocity is constant.

Horizontal component of velocity, u cos θ = 28 x Cos 30 = 24.25 m/s

At t = 2 sec, the horizontal component of velocity = 24.25 m/s

At t = 3 sec, the horizontal component of velocity = 24.25 m/s

i need the answer to this too but idk it ; -;