Answer:
it resembles volleyball because its "played as a variation of volleyball
Explanation:
Catchball — a easier version of volleyball in which the players catch and throw the ball rather than hit it. Volleyball — a game for two teams of six players, in which a large ball is hit by hand over a high net, the aim being to score points by making the ball reach the ground on the opponent's side of the court.
Answer:
Change in mechanical energy, 
Explanation:
It is given that,
Mass of the projectile, m = 12 kg
Speed of the projectile, v = 20 m/s
Maximum height, h = 18 m
Initially, the projectile have only kinetic energy. it is given by :
K = 2400 J
Finally, it have only potential energy. it is given by :
P = mgh
P =2116.8 J
The change in mechanical energy is given by :
So, the change in mechanical energy is 283.2 J. Hence, this is the required solution.
Answer:
It's due to the distance from either ends of strings origin...
Explanation:
As we know that waves behave moving in a flow from one side to another side and this gives a prospective of motion. Suppose a wave is pinched from the near one end of a guitar then due to the distortion created by the point of tie of strings the wave super imposes and moves with a velocity v and produces a wave frequency f. as we the pinching go down to the center the wave stabilizes itself to a stationary origin right at the center and the frequency then changes accordingly as moving down on the string.
We have the following equation for height:
h (t) = (1/2) * (a) * t ^ 2 + vo * t + h0
Where,
a: acceleration
vo: initial speed
h0: initial height.
The value of the acceleration is:
a = -g = -9.8 m / s ^ 2
For t = 0 we have:
h (0) = (1/2) * (a) * 0 ^ 2 + vo * 0 + h0
h (0) = h0
h0 = 0 (reference system equal to zero when the ball is hit).
For t = 5.8 we have:
h (5.8) = (1/2) * (- 9.8) * (5.8) ^ 2 + vo * (5.8) + 0
(1/2) * (- 9.8) * (5.8) ^ 2 + vo * (5.8) + 0 = 0
vo = (1/2) * (9.8) * (5.8)
vo = 28.42
Substituting values we have:
h (t) = (1/2) * (a) * t ^ 2 + vo * t + h0
h (t) = (1/2) * (- 9.8) * t ^ 2 + 28.42 * t + 0
Rewriting:
h (t) = -4.9 * t ^ 2 + 28.42 * t
The maximum height occurs when:
h '(t) = -9.8 * t + 28.42
-9.8 * t + 28.42 = 0
t = 28.42 / 9.8
t = 2.9 seconds.
Answer:
The ball was at maximum elevation when:
t = 2.9 seconds.
