Answer:
6.49707626552 m/s
4.45147808359 m
1.46212197842 s
Explanation:
t = Time taken
u = Initial velocity
v = Final velocity
s = Displacement
a = Acceleration
g = Acceleration due to gravity = 9.81 m/s²
[tex]v^2-u^2=2as\\\Rightarrow v=\sqrt{2as+u^2}\\\Rightarrow v=\sqrt{2\times 34.6\times 0.61+0^2}\\\Rightarrow v=6.49707626552\ m/s[/tex]
The speed with which Sam releases the ball is 6.49707626552 m/s
[tex]v^2-u^2=2as\\\Rightarrow s=\dfrac{v^2-u^2}{2g}\\\Rightarrow s=\dfrac{0^2-6.49707626552^2}{2\times -9.81}\\\Rightarrow s=2.15147808359\ m[/tex]
The ball reaches a height of 2.3+2.15147808359 = 4.45147808359 m
Distance to his head from the maximum height is 4.45147808359-1.83 = 2.62147808359 m
[tex]s=ut+\frac{1}{2}gt^2\\\Rightarrow 2.62147808359=0t+\frac{1}{2}\times 9.81\times t^2\\\Rightarrow t=\sqrt{\frac{2.62147808359\times 2}{9.81}}\\\Rightarrow t=0.731060989212\ s[/tex]
Time taken to go down to his head is 0.731060989212 seconds from the maximum height
Time taken from the moment the ball left his head height = 0.731060989212+0.731060989212 = 1.46212197842 s
