Answer:
a) Initial speed of the ball = 14.45 m/s
b) At height 6 m speed of ball = 9.55 m/s
c) Maximum height reached = 10.65 m
Explanation:
a) We have equation of motion
, where s is the displacement, u is the initial velocity, t is the time taken and a is the acceleration.
s = 6 m, t = 0.5 seconds, a = acceleration due to gravity value = -9.8![m/s^2](https://tex.z-dn.net/?f=m%2Fs%5E2)
Substituting
![6=u*0.5-\frac{1}{2} *9.8*0.5^2\\ \\ u=14.45m/s](https://tex.z-dn.net/?f=6%3Du%2A0.5-%5Cfrac%7B1%7D%7B2%7D%20%2A9.8%2A0.5%5E2%5C%5C%20%5C%5C%20u%3D14.45m%2Fs)
Initial speed of the ball = 14.45 m/s
b) We have equation of motion
, where v is the final velocity
s = 6 m, u = 14.45 m/s, a = -9.8![m/s^2](https://tex.z-dn.net/?f=m%2Fs%5E2)
Substituting
![v^2=14.45^2-2*9.8*6\\ \\ v=9.55m/s](https://tex.z-dn.net/?f=v%5E2%3D14.45%5E2-2%2A9.8%2A6%5C%5C%20%5C%5C%20v%3D9.55m%2Fs)
So at height 6 m speed of ball = 9.55 m/s
c) We have equation of motion
, where v is the final velocity
u = 14.45 m/s, v =0 , a = -9.8![m/s^2](https://tex.z-dn.net/?f=m%2Fs%5E2)
Substituting
![0^2=14.45^2-2*9.8*s\\ \\ s=10.65 m](https://tex.z-dn.net/?f=0%5E2%3D14.45%5E2-2%2A9.8%2As%5C%5C%20%5C%5C%20s%3D10.65%20m)
Maximum height reached = 10.65 m