A blue ball is thrown upward with an initial speed of 23 m/s, from a height of 0.5 meters above the ground. 2.8 seconds after the blue ball is thrown, a red ball is thrown down with an initial speed of 6.6 m/s from a height of 28.9 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.
1) What is the maximum height the blue ball reaches?
2) What is the height of the red ball 3.64 seconds after the blue ball is thrown?
3) How long after the blue ball is thrown are the two balls in the air at the same height?
4) Which statement is true about the blue ball after it has reached its maximum height and is falling back down?
a. The acceleration is positive and it is speeding up
b. The acceleration is negative and it is speeding up
c. The acceleration is positive and it is slowing down
d. The acceleration is negative and it is slowing down
Thank you!!!
just use your basic equations of motion:
h(t) = h0 + v0*t - 4.9t^2
v(t) = v0 - 9.8t
show what you get.
2. T = 3.64-2.8 = 0.84s.l
h = ho - (Vo*t + 0.5g*T^2).
h = 28.9 - (6.6*0.84 + 4.9*0.84^2),
4a. acceleration is positive and it is speeding up.
1) To find the maximum height the blue ball reaches, we can use the kinematic equation for vertical motion:
v^2 = u^2 + 2as
Where:
v is the final velocity (which is 0 when the ball reaches its maximum height)
u is the initial velocity (23 m/s)
a is the acceleration (-9.81 m/s^2, considering downward motion)
s is the displacement (the height we want to find)
Rearranging the equation, we get:
0 = (23)^2 + 2(-9.81)s
Simplifying:
0 = 529 - 19.62s
19.62s = 529
s = 529 / 19.62
s ≈ 26.99 meters
Therefore, the maximum height the blue ball reaches is approximately 26.99 meters.
2) To find the height of the red ball 3.64 seconds after the blue ball is thrown, we can use the kinematic equation for vertical motion again. The initial height is given as 28.9 meters, the initial velocity is -6.6 m/s (negative because it is thrown downward), and the acceleration is -9.81 m/s^2.
Using the equation:
s = ut + (1/2)at^2
Where:
s is the displacement (the height we want to find)
u is the initial velocity (-6.6 m/s)
t is the time in seconds (3.64 seconds)
a is the acceleration (-9.81 m/s^2)
Plugging in the values:
s = (-6.6)(3.64) + (1/2)(-9.81)(3.64)^2
Simplifying:
s = -23.98 + (1/2)(-9.81)(13.25)
s = -23.98 - (65.01)
s = -88.99
Therefore, the height of the red ball 3.64 seconds after the blue ball is thrown is approximately -88.99 meters. Since the height cannot be negative in this scenario, we can conclude that the red ball has hit the ground.
3) To find the time after the blue ball is thrown when the two balls are at the same height, we can set up equations for their heights at any given time. For the blue ball, the equation is:
s_b = 0.5 + 23t + (1/2)(-9.81)t^2
For the red ball, the equation is:
s_r = 28.9 + (-6.6)t + (1/2)(-9.81)t^2
We need to find the time (t) when s_b = s_r. Setting the two equations equal to each other:
0.5 + 23t + (1/2)(-9.81)t^2 = 28.9 + (-6.6)t + (1/2)(-9.81)t^2
Rearranging and simplifying:
23t + 0.5 = -6.6t + 28.9
29.6t = 28.4
t = 28.4 / 29.6
t ≈ 0.96 seconds
Therefore, the two balls are at the same height approximately 0.96 seconds after the blue ball is thrown.
4) After the blue ball has reached its maximum height and is falling back down, the acceleration is negative and it is slowing down.
The correct answer is d. The acceleration is negative and it is slowing down.
To find the answers to the given questions, we can use the kinematic equations of motion. These equations relate the displacement, velocity, acceleration, and time for an object undergoing constant acceleration.
Let's analyze each question step by step:
1) To calculate the maximum height the blue ball reaches, we first need to determine the time it takes for the blue ball to reach its maximum height. We can use the equation:
vf = vi + a*t
where vf is the final velocity, vi is the initial velocity, a is the acceleration, and t is the time.
Since the ball is thrown upward, its final velocity at the maximum height will be 0 m/s. The initial velocity is 23 m/s, and the acceleration is -9.81 m/s² (negative because it opposes the upward motion). Solving for t:
0 = 23 + (-9.81)*t
t = 2.35 seconds.
Now, we can calculate the maximum height using the equation:
d = vi*t + (1/2)*a*t^2
where d is the displacement.
d = 23*2.35 + (1/2)*(-9.81)*(2.35)^2
d = 27.022 m
Therefore, the maximum height the blue ball reaches is approximately 27.022 meters.
2) To calculate the height of the red ball 3.64 seconds after the blue ball is thrown, we can use the same equation:
d = vi*t + (1/2)*a*t^2
The initial velocity of the red ball is -6.6 m/s (negative because it is thrown downwards), time t = 3.64 s, and acceleration a = 9.81 m/s² (positive because it is in the downward direction).
d = -6.6*3.64 + (1/2)*9.81*(3.64)^2
d = -71.168 m
Therefore, the height of the red ball 3.64 seconds after the blue ball is thrown is approximately -71.168 meters. A negative value indicates that the ball is below the initial height.
3) To find the time when both balls are at the same height, we need to set the displacement equations for both balls equal to each other:
Blue ball displacement = red ball displacement
Initial height + blue ball displacement = Initial height + red ball displacement
0.5 + 23*t - (1/2)*9.81*t^2 = 28.9 - 6.6*t - (1/2)*9.81*t^2
Simplifying the equation:
23*t = 28.4 - 6.6*t
29.6*t = 28.4
t = 0.96 seconds
Therefore, the blue ball and red ball are at the same height approximately 0.96 seconds after the blue ball is thrown.
4) After the blue ball has reached its maximum height and is falling back down, the acceleration is negative due to the force of gravity. In this scenario, the statement "d. The acceleration is negative and it is slowing down" is true.
I hope this explanation helps you understand how to solve the given questions!