Answer:
The maximum height of ball 2 is 4 times that of ball 1
Explanation:
We can find the maximum height of each ball by using the following suvat equation:
[tex]v^2-u^2=2as[/tex]
where
v is the final velocity
u is the initial velocity
[tex]a=g=-9.8 m/s^2[/tex] is the acceleration of gravity (we take upward as positive direction)
s is the displacement
At the maximum height, s = h and v = 0 (the final velocity is zero), so re-arranging the equation:
[tex]h=\frac{-u^2}{2g}[/tex]
The first ball is thrown with initial velocity [tex]u_1[/tex], so it reaches a maximum height of
[tex]h_1 = -\frac{u_1^2}{2g}[/tex] (the quantity will be positive, since g is negative)
The second ball is thrown with initial velocity
[tex]u_2 = u_1[/tex]
so it will reach a maximum height of
[tex]h_2 = - \frac{u_2^2}{2g}=-\frac{(2u_1)^2}{2g}=4(-\frac{u_1^2}{2g}) = 4h_1[/tex]
So, its maximum height will be 4 times the maximum height reached by ball 1.
