A ball is thrown upward from the ground with an initial speed of 25 m/s; at the same instant, a ball is dropped from rest from a building 15 m high. After how long will the balls be at the same height?
To find out after how long the balls will be at the same height, we need to calculate the time it takes for each ball to reach that height.
First, let's consider the ball that is thrown upward. We know that the initial velocity (u) of the ball is 25 m/s, and the acceleration due to gravity (a) is -9.8 m/s^2 (negative because it's acting in the opposite direction to the motion). The height (h) at which the balls will be at the same height is 15 meters.
We can use the equation for displacement: h = ut + (1/2)at^2, where t is the time taken.
For the ball thrown upward:
h = 15 m
u = 25 m/s
a = -9.8 m/s^2
Using the above equation, we can rearrange it to solve for time (t):
15 = 25t + (1/2)(-9.8)t^2
Simplifying the equation:
15 = 25t - 4.9t^2
This equation is a quadratic equation. We can solve it by setting it equal to zero and using either factoring, completing the square, or using the quadratic formula. Let's use the quadratic formula to find the value of t when the ball thrown upward is at the same height:
t = (-b ± √(b^2 - 4ac))/(2a)
Here, a = -4.9, b = 25, and c = -15.
Plugging in the values:
t = (-25 ± √(25^2 - 4*(-4.9)*(-15)))/(2*(-4.9))
Simplifying:
t = (-25 ± √(625 - 294))/(2*(-4.9))
t = (-25 ± √(331))/(2*(-4.9))
Now, solving for t:
t ≈ (-25 ± √(331))/(-9.8)
Calculating the value:
t ≈ (-25 ± √331)/(-9.8)
Using a calculator, we find two possible solutions:
t ≈ 1.94 seconds or t ≈ 3.04 seconds
Therefore, after approximately 1.94 seconds and 3.04 seconds, the ball thrown upward and the ball dropped from the building will be at the same height.