Physics Volleyball Problem

Question

A volleyball is slammed from a height of 2.8 m above the ground, at an angle of 76 degrees as measured counterclockwise from the -y axis, toward an opponent. It misses the opponent and strikes the ground 8.0 m away and scores a point.

a) How long was the volleyball in the air?

b) What was the initial speed of the volleyball?

Answer 1

To solve this problem, we can use the kinematic equations of motion for projectile motion. We can break the initial velocity of the volleyball into its x- and y-components:

vx = v0 cos(76°) vy = v0 sin(76°)

where v0 is the initial speed of the volleyball.

a) To find how long the volleyball was in the air, we can use the fact that the time of flight, tf, for a projectile launched from a height h0 and landing at the same height is given by:

tf = 2 * vy / g

where g is the acceleration due to gravity (approximately 9.81 m/s^2).

Substituting the given values, we get:

tf = 2 * v0 sin(76°) / g

To find v0, we can use the fact that the horizontal range, R, for a projectile launched at an angle θ from a height h0 and landing at the same height is given by:

R = v0^2 * sin(2θ) / g

Substituting the given values, we get:

8.0 m = v0^2 * sin(152°) / g

Solving for v0, we get:

v0 = √(8.0 m * g / sin(152°)) ≈ 7.5 m/s

Now we can substitute this value of v0 into the equation for tf:

tf = 2 * v0 sin(76°) / g ≈ 1.1 s

Therefore, the volleyball was in the air for approximately 1.1 seconds.

b) The initial speed of the volleyball is approximately 7.5 m/s.