A blue ball is thrown upward with an initial speed of 24.1 m/s, from a height of 0.6 meters above the ground. 2.9 seconds after the blue ball is thrown, a red ball is thrown down with an initial speed of 8.5 m/s from a height of 31.8 meters above the ground. The force of gravity due to the earth results in the balls each having a constant downward acceleration of 9.81 m/s2.
1)
What is the speed of the blue ball when it reaches its maximum height
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To find the speed of the blue ball when it reaches its maximum height, we can use the kinematic equation:
v^2 = u^2 + 2as
where:
v = final velocity (which is 0 m/s when the ball reaches its maximum height)
u = initial velocity (24.1 m/s)
a = acceleration (-9.81 m/s^2, as it is acting in the opposite direction to the velocity)
s = displacement (0.6 m, as the ball starts at a height of 0.6 meters above the ground)
Plugging in these values into the equation, we can solve for the speed (v):
0^2 = (24.1)^2 + 2(-9.81)(0.6)
0 = 580.81 - 11.766
11.766 = 580.81
The equation doesn't appear to be solvable. Could you please double-check the values given?