a.) using the free fall height equation ... -16 t^2 + 80 = -16 t^2 + 40 t
b.) plug the t from a.) into either expression ... -16 t^2 + 80 or -16 t^2 + 40 t
b.) plug the t from a.) into either expression ... -16 t^2 + 80 or -16 t^2 + 40 t
a) When do the balls pass each other:
The motion of the ball dropped from the top of the tower (Ball A) can be described by the equation of motion:
y1 = y0 + v0t + (1/2)gt^2
Where:
y1 is the final position (height) of the ball A
y0 is the initial position (height) of the ball A (80 ft)
v0 is the initial velocity of the ball A (0 ft/s, as it is dropped)
g is the acceleration due to gravity (-32 ft/s^2, considering upward as positive direction)
t is the time
Similarly, the motion of the ball thrown upward from the ground (Ball B) can be described by the equation of motion:
y2 = y0 + v0t + (1/2)gt^2
Where:
y2 is the final position (height) of the ball B
y0 is the initial position (height) of the ball B (0 ft)
v0 is the initial velocity of the ball B (40 ft/s, as it is thrown upward)
g is the acceleration due to gravity (-32 ft/s^2, considering upward as positive direction)
t is the time
To find when they pass each other, we need to find the time (t) when y1 = y2.
Setting y1 = y2, we have:
y0 + v0*t + (1/2)g*t^2 = y0 + v0*t + (1/2)g*t^2
Simplifying the equation:
80 + 0*t + (1/2)(-32)*t^2 = 0 + 40*t + (1/2)(-32)*t^2
Simplifying further:
80 = 40t
Dividing both sides by 40, we find:
t = 2 seconds
Therefore, the two balls pass each other after 2 seconds.
b) Where do they pass each other:
To find where they pass each other, we need to find the position (height) at that time (t = 2).
Using the equation of motion for Ball A:
y1 = y0 + v0*t + (1/2)g*t^2
Substituting the values:
y1 = 80 + 0*2 + (1/2)(-32)*(2)^2
Calculating:
y1 = 80 - 32
y1 = 48 ft
Therefore, the two balls pass each other at a height of 48 ft above the ground.