A blue ball is thrown upward with an initial speed of 19.6 m/s, from a height of 0.5 meters above the ground. 2.4 seconds after the blue ball is thrown, a red ball is thrown down with an initial speed of 7.9 m/s from a height of 22.1 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/s^2.

Question

A blue ball is thrown upward with an initial speed of 19.6 m/s, from a height of 0.5 meters above the ground. 2.4 seconds after the blue ball is thrown, a red ball is thrown down with an initial speed of 7.9 m/s from a height of 22.1 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/s^2.

One question..
How long after the blue ball is thrown are the two balls in the air at the same height?..and don't say it is .786 because that's only if they were released at the same time. Thanks

Answer 1

heightblue=.5+19.6t-4.9t^2

heightred=+22.1 -7.9(t-2.4)-4.9(t-2.4)^2

set these heights equal, solve for t. Have fun.

Answer 2

Hey bobpursley! ITS 3.09!! YAYY thanks lot!

Answer 3

where did you get the 4.9

Answer 4

To find out how long after the blue ball is thrown the two balls are at the same height, we need to calculate the time it takes for each ball to reach that height.

Let's start with the blue ball. The initial velocity is 19.6 m/s, and the initial height is 0.5 meters above the ground. The acceleration is constant at -9.81 m/s^2 (negative because it is acting in the opposite direction to the initial velocity).

We can use the kinematic equation to find the time it takes for the blue ball to reach the same height as the red ball:

h = h0 + v0*t + (1/2)*a*t^2

Here, h is the height reached by the blue ball (the same height as the red ball), h0 is the initial height (0.5 meters), v0 is the initial velocity (19.6 m/s), a is the acceleration (-9.81 m/s^2), and t is the time.

Rearranging the equation to solve for t:

h - h0 = v0*t + (1/2)*a*t^2
0 = v0*t + (1/2)*a*t^2
(1/2)*a*t^2 + v0*t = 0
t*( (1/2)*a*t + v0 ) = 0

Now, we have two possible solutions: t = 0 or ((1/2)*a*t + v0) = 0.

Since t = 0 represents the initial time when the blue ball is thrown (and we are interested in the time afterward), we can ignore that solution.

Now, we can solve for t using the second equation:

((1/2)*a*t + v0) = 0
(1/2)*(-9.81)*t + 19.6 = 0
-4.905*t + 19.6 = 0
-4.905*t = -19.6
t = -19.6 / -4.905
t = 4 seconds (rounded to the nearest whole number)

Therefore, 4 seconds after the blue ball is thrown, the two balls are at the same height.