To1. PE = mgh
PE potential energy
m mass
g 9.81
h height
P = PE / t
P power
t time
Answer:
Omg tysm you are too kind
Explanation:
Answer:
28.73 m from the base of the cliff collide happen
Explanation:
Equation for ball dropped from 50 cliff is reaches distance x in t seconds isi given by
![x=ut+\frac{1}{2} at^2\\\\x= 0 \times t+\frac{1}{2} gt^2 =\frac{1}{2} gt^2......(1)](https://tex.z-dn.net/?f=x%3Dut%2B%5Cfrac%7B1%7D%7B2%7D%20at%5E2%5C%5C%5C%5Cx%3D%200%20%5Ctimes%20t%2B%5Cfrac%7B1%7D%7B2%7D%20gt%5E2%20%3D%5Cfrac%7B1%7D%7B2%7D%20gt%5E2......%281%29)
stone is thrown up from the bottom with speed u=24 m/s . it reaches distance y when stone collide with ball.(g is negative here)
![y=24t-\frac{1}{2} gt^2.............(2)](https://tex.z-dn.net/?f=y%3D24t-%5Cfrac%7B1%7D%7B2%7D%20gt%5E2.............%282%29)
we know that total distance traveled by ball and stone is 50 m
![x+y=50 m](https://tex.z-dn.net/?f=x%2By%3D50%20m)
adding equation 1 and 2, we get time t
![x+y=\frac{1}{2} gt^2+24t-\frac{1}{2}gt^2\\50=24t\\t=2.083 s](https://tex.z-dn.net/?f=x%2By%3D%5Cfrac%7B1%7D%7B2%7D%20gt%5E2%2B24t-%5Cfrac%7B1%7D%7B2%7Dgt%5E2%5C%5C50%3D24t%5C%5Ct%3D2.083%20s)
substitute this time in equation 2, we can get the required distance where they collide
![y=24t-\frac{1}{2} gt^2\\y=24\times 2.083-\frac{1}{2}\times 9.8\times 2.083^2\\y=28.73 m](https://tex.z-dn.net/?f=y%3D24t-%5Cfrac%7B1%7D%7B2%7D%20gt%5E2%5C%5Cy%3D24%5Ctimes%202.083-%5Cfrac%7B1%7D%7B2%7D%5Ctimes%20%209.8%5Ctimes%202.083%5E2%5C%5Cy%3D28.73%20m)
28.73 m from the base of the cliff collide happen