Answer:
a) 256 ft
b) 32 ft/s
c) -32 ft/s
Explanation:
t = Time taken
u = Initial velocity
v = Final velocity
s = Displacement
a = Acceleration due to gravity = 32 ft/s²
[tex]v^2-u^2=2as\\\Rightarrow s=\frac{v^2-u^2}{2a}\\\Rightarrow s=\frac{0^2-128^2}{2\times -32}\\\Rightarrow s=256\ ft[/tex]
Maximum height = 256 ft
When s = 240 ft
[tex]v^2-u^2=2as\\\Rightarrow v=\sqrt{2as+u^2}\\\Rightarrow v=\sqrt{2\times -32\times 240+128^2}\\\Rightarrow v=32\ ft/s[/tex]
Speed of the ball at 240 ft is 32 ft/s while going up
Now this will be the velocity of the ball when it reaches 240 ft, which will be considered as the initial velocity
[tex]v=u+at\\\Rightarrow 0=32-32t\\\Rightarrow t=\frac{-32}{-32}=1\ s[/tex]
Now, initial velocity will be considered as zero
[tex]v^2-u^2=2as\\\Rightarrow v=\sqrt{2as+u^2}\\\Rightarrow v=\sqrt{2\times 32\times 16+0^2}\\\Rightarrow v=32\ ft/s[/tex]
Speed of the ball at 240 ft is -32 ft/s while going down
