The motion of the tennis ball on the vertical axis is an uniformly accelerated motion, with deceleration of
![g=-9.81 m/s^2](https://tex.z-dn.net/?f=g%3D-9.81%20m%2Fs%5E2)
(gravitational acceleration).
The component of the velocity on the y-axis is given by the following law:
![v_y(t) = v_{y0}+gt](https://tex.z-dn.net/?f=v_y%28t%29%20%3D%20v_%7By0%7D%2Bgt)
At the time t=0.5 s, the ball reaches its maximum height, and when this happens, the vertical velocity is zero (because it is a parabolic motion):
![v_y(0.5 s)=0](https://tex.z-dn.net/?f=v_y%280.5%20s%29%3D0)
. Substituing into the previous equation, we find the initial value of the vertical component of the velocity:
![v_{y0}=-gt=-(-9.81 m/s^2)(0.5 s)=4.9 m/s](https://tex.z-dn.net/?f=v_%7By0%7D%3D-gt%3D-%28-9.81%20m%2Fs%5E2%29%280.5%20s%29%3D4.9%20m%2Fs)
However, this is not the final answer. In fact, the ball starts its trajectory with an angle of
![30^{\circ}](https://tex.z-dn.net/?f=30%5E%7B%5Ccirc%7D)
. This means that the vertical component of the initial velocity is
![v_{y0}=v_0 sin 30^{\circ}](https://tex.z-dn.net/?f=v_%7By0%7D%3Dv_0%20sin%2030%5E%7B%5Ccirc%7D)
We found before
![v_{0y}=4.9 m/s](https://tex.z-dn.net/?f=v_%7B0y%7D%3D4.9%20m%2Fs)
, so we can substitute to find
![v_0](https://tex.z-dn.net/?f=v_0)
, the initial speed of the ball:
Answer:2726.41 W
Explanation:
Consider the voltage and current in the form of
![v\left ( t\right )=V_mcos\omega t](https://tex.z-dn.net/?f=v%5Cleft%20%28%20t%5Cright%20%29%3DV_mcos%5Comega%20t)
![i\left ( t\right )=I_mcos\omega t](https://tex.z-dn.net/?f=i%5Cleft%20%28%20t%5Cright%20%29%3DI_mcos%5Comega%20t)
![p\left ( t\right )=\left ( V_mI_m\right )\left ( cos\omega t\right )^2](https://tex.z-dn.net/?f=p%5Cleft%20%28%20t%5Cright%20%29%3D%5Cleft%20%28%20V_mI_m%5Cright%20%29%5Cleft%20%28%20cos%5Comega%20t%5Cright%20%29%5E2)
![p\left ( t\right )=\left ( V_mI_m\right )\left [ \frac{1+cos2\omega t}{2} \right ]](https://tex.z-dn.net/?f=p%5Cleft%20%28%20t%5Cright%20%29%3D%5Cleft%20%28%20V_mI_m%5Cright%20%29%5Cleft%20%5B%20%5Cfrac%7B1%2Bcos2%5Comega%20t%7D%7B2%7D%20%5Cright%20%5D)
![p\left ( t\right )=\frac{ V_mI_m}{2}\left ( 1+cos2\omega t\right )](https://tex.z-dn.net/?f=p%5Cleft%20%28%20t%5Cright%20%29%3D%5Cfrac%7B%20V_mI_m%7D%7B2%7D%5Cleft%20%28%201%2Bcos2%5Comega%20t%5Cright%20%29)
![P_{avg}=\frac{ V_mI_m}{2}](https://tex.z-dn.net/?f=P_%7Bavg%7D%3D%5Cfrac%7B%20V_mI_m%7D%7B2%7D)
![P_{avg}=\frac{V_m^2}{2r}](https://tex.z-dn.net/?f=P_%7Bavg%7D%3D%5Cfrac%7BV_m%5E2%7D%7B2r%7D)
![P_{avg}=\frac{170^2}{5.3\times 2}=2726.41W](https://tex.z-dn.net/?f=P_%7Bavg%7D%3D%5Cfrac%7B170%5E2%7D%7B5.3%5Ctimes%202%7D%3D2726.41W)
Answer:
The answer is radio waves
There are 1,000,000 micro seconds in one second so multiple 136.8 by 1000000 and you'll get 136,800,000 Tera calculations per second.