A ball is thrown upward from the ground with an initial speed of 23.1 m/s; at the same instant, a ball is dropped from a building 15.0 m high. After how long will the balls be at the same height?
Find the t value that satisfies this equation:
Y1(t) = Y2(t)
23.1 t - 4.9 t^2 = 15 - 4.9t^2
23.1 t = 15
t = 0.649 s
To determine the time at which both balls are at the same height, we can use the basic principles of projectile motion.
First, let's analyze the motion of the ball thrown upward. We'll use the equation:
h = h0 + v0t - 0.5gt^2
Where:
h = final height
h0 = initial height (which is 0 for a ball thrown upward from the ground)
v0 = initial velocity (23.1 m/s)
g = acceleration due to gravity (-9.8 m/s^2)
t = time
Since we want to find the time when the ball is at the same height, we'll set h equal to the height of the building (15.0 m):
15.0 = 0 + (23.1)t - 0.5(9.8)t^2
Now, let's analyze the motion of the ball dropped from the building. In this case, the initial height is 15.0 m, and the initial velocity is 0. The equation for this motion is:
h = h0 + v0t - 0.5gt^2
We'll use the same value of g and set h equal to 15.0 m:
15.0 = 15.0 + 0 - 0.5(9.8)t^2
Now we have two equations with the same unknown, t. We can solve them simultaneously to find the value of t when the balls are at the same height.
By rearranging the equation for the ball thrown upward, we get:
15.0 = 23.1t - 0.5(9.8)t^2
Simplifying, we have:
0.5(9.8)t^2 - 23.1t + 15.0 = 0
Now, we can use the quadratic formula to solve for t:
t = (-b ± √(b^2 - 4ac)) / (2a)
In our case:
a = 0.5(9.8) = 4.9
b = -23.1
c = 15.0
Plugging those values into the quadratic formula, we get:
t = (-(-23.1) ± √((-23.1)^2 - 4(0.5(9.8))(15.0))) / (2(0.5(9.8)))
Simplifying further:
t = (23.1 ± √(534.48 - 294.0)) / 4.9
t = (23.1 ± √240.48) / 4.9
Now, we can calculate the two possible values of t:
t1 = (23.1 + √240.48) / 4.9
t2 = (23.1 - √240.48) / 4.9
Calculating the values, we find:
t1 ≈ 2.997 seconds
t2 ≈ 0.333 seconds
Since time cannot be negative, we discard t2 as an extraneous solution. The time at which both balls are at the same height is approximately 2.997 seconds.