The first ball reaches the ground first
Explanation:
We can solve the problem by using suvat equations, since the motion of both balls is a free fall motion (with constant acceleration, [tex]g=9.8 m/s^2[/tex], towards the ground).
The equation of motion that represents the y-position of the first ball at time t is
[tex]y_1 = u_1 t + \frac{1}{2}at^2[/tex]
where
[tex]u_1 = 41.67 m/s[/tex] is the initial vertical velocity of the ball
[tex]a=-g=-9.8 m/s^2[/tex] is the acceleration (downward, therefore negative)
Substituting [tex]y_1 = 0[/tex] and solving for t, we find the corresponding time at which the ball reaches the ground:
[tex]0=u_1 t + \frac{1}{2}at^2\\0=t(u_1 + \frac{1}{2}at)[/tex]
The two solutions are:
t = 0 (starting moment)
[tex]u_1 + \frac{1}{2}at=0\\t=-\frac{2u_1}{a}=-\frac{2(41.67)}{-9.8}=8.5 s[/tex]
So, the first ball reaches the ground after 8.5 s.
Similarly, for the second ball
[tex]y_2 = u_2 t + \frac{1}{2}at^2[/tex]
where
[tex]u_2 = 55.56 m/s[/tex] is the initial vertical velocity of the ball
[tex]a=-g=-9.8 m/s^2[/tex] is the acceleration (downward, therefore negative)
Substituting [tex]y_2 = 0[/tex] and solving for t, we find the corresponding time at which the ball reaches the ground:
[tex]0=u_2 t + \frac{1}{2}at^2\\0=t(u_2 + \frac{1}{2}at)[/tex]
The two solutions are:
t = 0 (starting moment)
[tex]u_2 + \frac{1}{2}at=0\\t=-\frac{2u_2}{a}=-\frac{2(55.56)}{-9.8}=11.3 s[/tex]
So, the second ball reaches the ground after 11.3 s. However, the ball has been thrown 2 seconds after the first ball, so the actual time is
t = 11.3 + 2 = 13.3 s
This means that the first ball reaches the ground first.
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