A ball is thrown upward from the ground with an initial speed of 23.2 m/s; at the same instant, another ball is dropped from a building 20 m high. After how long will the balls be at the same height?

Given:

Let the first ball have height x, velocity u.
x(0) = 0[m]
u(0) = 23.2[m/s]
x(t) = x(0) + u(0) t + (g/2) t^2
.: x(t) = 23.2 t - 4.9 t^2

Let the second have height y, velocity v.
y(0) = 20[m]
v(0) = 0[m/s]
y(t) = y(0) + v(0) t + (g/2) t^2
.: y(t) = 20 - 4.9 t^2

Where: g = -9.8[m/s^2]

Find: t such that x(t)=y(t)

I set the two position equations together and got 0.86 s as my time. Thanks!

To determine the time when both balls are at the same height, we need to find the time it takes for each ball to reach that height.

Let's start with the ball thrown upward. We can use the kinematic equation:

š‘¦ = š‘£ā‚€š‘” - 0.5š‘”š‘”Ā²

where š‘¦ is the height, š‘£ā‚€ is the initial velocity, š‘” is the acceleration due to gravity, and š‘” is the time.

For the ball thrown upward:
š‘¦ā‚ = š‘£ā‚€š‘” - 0.5š‘”š‘”Ā²

Since the ball is thrown with an initial speed of 23.2 m/s, we have:
š‘£ā‚€ā‚ = 23.2 m/s

The acceleration due to gravity is approximately 9.8 m/sĀ².

For the ball dropped vertically, we can use a simplified version of the equation, as it has no initial velocity:
š‘¦ā‚‚ = 0.5š‘”š‘”Ā²

Since the ball is dropped from a height of 20 m, we have:
š‘¦ā‚‚ = 20 m

Now, we can set š‘¦ā‚ equal to š‘¦ā‚‚ and solve for š‘”:

š‘£ā‚€ā‚š‘” - 0.5š‘”š‘”Ā² = 0.5š‘”š‘”Ā²

Simplifying the equation leads to:
š‘£ā‚€ā‚š‘” = š‘”š‘”Ā²

Dividing both sides by š‘” gives us:
š‘£ā‚€ā‚ = š‘”š‘”

Plugging in the values:
23.2 m/s = 9.8 m/sĀ² Ɨ š‘”

Now, solve for š‘”:
š‘” = 23.2 m/s / 9.8 m/sĀ² ā‰ˆ 2.37 s

Therefore, after approximately 2.37 seconds, both balls will be at the same height.