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 [tex]s=ut+\frac{1}{2} at^2[/tex], 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[tex]m/s^2[/tex]
Substituting
[tex]6=u*0.5-\frac{1}{2} *9.8*0.5^2\\ \\ u=14.45m/s[/tex]
Initial speed of the ball = 14.45 m/s
b) We have equation of motion [tex]v^2=u^2+2as[/tex], where v is the final velocity
s = 6 m, u = 14.45 m/s, a = -9.8[tex]m/s^2[/tex]
Substituting
[tex]v^2=14.45^2-2*9.8*6\\ \\ v=9.55m/s[/tex]
So at height 6 m speed of ball = 9.55 m/s
c) We have equation of motion [tex]v^2=u^2+2as[/tex], where v is the final velocity
u = 14.45 m/s, v =0 , a = -9.8[tex]m/s^2[/tex]
Substituting
[tex]0^2=14.45^2-2*9.8*s\\ \\ s=10.65 m[/tex]
Maximum height reached = 10.65 m
