Abasketball player is trying to make a half-court jump shot, and releasing the ball at the height of the basket. assuming that the ball is launched at 51.0â°, 16.0 m from the basket, what speed must the player give the ball?

The player must give a speed of 12.66 m/s to the ball.

Explanation:

Since, the ball is released at the same height as the basket. Therefore, this is a complete projectile motion, with following parameters:

Horizontal Range = R = 16 m

Launch Angle = θ = 51°

Launch Speed = V = ?

The formula for horizontal range of projectile is:

R = V² Sin 2θ/g

V² = Rg / Sin 2θ

V² = (16 m)(9.8 m/s²) / Sin 2(51°)

V = √160.3 m²/s²

V = 12.66 m/s

a small object of mass 3.66 g and charge -17.6 µc is suspended motionless above the ground when immersed in a uniform electric field perpendicular to the ground. what are the magnitude and direction of the electric field?

c.the charges if the proton and electron and gravity