a ball is dropped from the top of the tower 80 ft.high at the same instant that a second ball is thrown upward from the ground with an initial velocity of 40 ft/s.a.)When do they pass each other and b.)where do the pass each other.
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
To find when and where the two balls pass each other, we need to determine the time and position at which their heights become equal.
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.