A red ball is thrown down with an initial speed of 1.4 m/s from a height of 28 meters above the ground. Then, 0.6 seconds after the red ball is thrown, a blue ball is thrown upward with an initial speed of 24.9 m/s, from a height of 0.9 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.
What is the height of the blue ball 1.9 seconds after the red ball is thrown
Well, it seems like the red ball just couldn't wait for the blue ball to join the party! Anyway, let's calculate the height of the blue ball.
To do this, we'll use the equation for vertical displacement, which is given by:
h = u*t + (1/2)*a*t^2
Where:
- h is the height or displacement,
- u is the initial velocity,
- a is the acceleration, and
- t is the time.
For the red ball:
- u = 1.4 m/s (downwards),
- a = 9.81 m/s^2 (downwards), and
- t = 0.6 seconds.
Using the equation, we find the vertical displacement of the red ball:
h_red = (1.4 m/s)*(0.6 s) + (1/2)*(9.81 m/s^2)*(0.6 s)^2
Now, let's calculate the height of the blue ball at 1.9 seconds after the red ball is thrown. For the blue ball:
- u = 24.9 m/s (upwards),
- a = -9.81 m/s^2 (downwards, since it opposes the ball's upward motion), and
- t = 1.9 seconds.
Using the same equation, we get:
h_blue = (24.9 m/s)*(1.9 s) + (1/2)*(-9.81 m/s^2)*(1.9 s)^2
Now let me get my calculator to compute those values...
To find the height of the blue ball 1.9 seconds after the red ball is thrown, we need to consider the motion of both balls separately and calculate their respective positions.
Let's start by finding the position of the red ball after 1.9 seconds. We know that the initial speed of the red ball is 1.4 m/s, and the acceleration due to gravity is -9.81 m/s^2 (negative because it's acting downwards). We can use the following kinematic equation to calculate its position:
h_red = h_red0 + v_red0 * t + 0.5 * a_red * t^2
where:
h_red is the height of the red ball after time t
h_red0 is the initial height of the red ball
v_red0 is the initial velocity of the red ball
a_red is the acceleration of the red ball
t is the time in seconds
Plugging in the given values:
h_red0 = 28 m
v_red0 = 1.4 m/s
a_red = -9.81 m/s^2
t = 1.9 s
h_red = 28 + 1.4 * 1.9 + 0.5 * (-9.81) * (1.9)^2
Calculating this equation gives us the height of the red ball after 1.9 seconds. Now, let's move on to the blue ball.
The blue ball is thrown upward, so its initial velocity is positive. We'll use a similar equation to find its position after 1.9 seconds:
h_blue = h_blue0 + v_blue0 * t + 0.5 * a_blue * t^2
where:
h_blue is the height of the blue ball after time t
h_blue0 is the initial height of the blue ball
v_blue0 is the initial velocity of the blue ball
a_blue is the acceleration of the blue ball
t is the time in seconds
Plugging in the given values:
h_blue0 = 0.9 m
v_blue0 = 24.9 m/s
a_blue = -9.81 m/s^2
t = 1.9 s
h_blue = 0.9 + 24.9 * 1.9 + 0.5 * (-9.81) * (1.9)^2
Calculating this equation gives us the height of the blue ball after 1.9 seconds.
Note: Since the blue ball is thrown upwards, its acceleration is still -9.81 m/s^2, but its initial velocity is positive.
By solving these equations, we can find the heights of both balls after 1.9 seconds, which will allow us to determine the height of the blue ball.