A volleyball is launched straight up at an initial velocity of 2.3 m/s. It is then spiked by a taller player 0.8 meters above the launch height. How high did it rise, and how long did the volleyball remain in the air between the players?
V^2 = Vo^2 + 2g*h = 0 @ max ht.
h max = -(Vo^2)/2g = -(2.3^2)/-19.6 = 0.270 m.
Since the max ht. is only 0.27 m, the
ball cannot be spiked at o.8 m. Please
check the numbers for accuracy.
To find out how high the volleyball rose and how long it remained in the air, we can use the equations of motion for vertical projectile motion.
Let's break down the information given:
Initial velocity (v0) = 2.3 m/s (upward)
Height above launch point (h) = 0.8 meters
To determine the maximum height reached by the volleyball (h_max), we can use the following equation:
h_max = [(v0^2) / (2 * g)] + h
where g is the acceleration due to gravity (approximately 9.8 m/s^2).
Plugging in the given values, we get:
h_max = [(2.3^2) / (2 * 9.8)] + 0.8 = 0.326 meters
Therefore, the volleyball reached a maximum height of approximately 0.326 meters.
Now, to find the total time of flight (T) of the volleyball, we can use the equation:
T = 2 * v0 / g
Substituting the values, we get:
T = 2 * 2.3 / 9.8 = 0.469 seconds
Therefore, the volleyball remained in the air for approximately 0.469 seconds.
In summary, the volleyball rose to a height of approximately 0.326 meters and remained in the air for approximately 0.469 seconds.