2 balls dropped from a building

[ronny45]

Homework Statement

A ball is thrown straight up from the edge of the roof of a building. A second ball is dropped from the roof a time of 1.00s later. You may ignore air resistance.
If the height of the building is 21.0m, what must the initial speed be of the first ball if both are to hit the ground at the same time?

Homework Equations

v=u+at
v^2=u^2+2as
2s=ut+1/2 at^2

The Attempt at a Solution

For the second ball,
u=0
a=9.81m/s^2
s=21m
v=?
...v comes out to be 20.298m. Then i found t to be 2.069s. Assuming the first ball was in the air 1 second longer, its time is 3.069.

However, I' not sure how to proceed with the equations for the first ball, since I don't know the maximum height reached by it, though v is 0 at this point.

 

[gknowels]

since you know the time, acceleration, initial and final positions, you should be able to use x = x_0 + v_0 t + (1/2) a t^2 to find the answer.

 

[ogb p]

I would approach this problem by first identifying the variables and equations that are relevant to the situation. In this case, the relevant variables are initial velocity (u), final velocity (v), acceleration (a), and time (t). The equations that can be used to solve this problem are the kinematic equations for motion with constant acceleration.

Using the equation v=u+at, we can calculate the final velocity of the second ball when it hits the ground. As the ball is dropped from rest, its initial velocity is 0, so we have v=at. Plugging in the values, we get v=9.81m/s^2 * 1.00s = 9.81m/s.

Next, we can use the equation 2s=ut+1/2at^2 to solve for the initial velocity of the first ball. We know that the height (s) of the building is 21.0m and the time (t) is 3.069s, since the first ball is in the air 1 second longer than the second ball. Plugging in these values, we get 2(21.0m)=u(3.069s)+1/2(9.81m/s^2)(3.069s)^2. Solving for u, we get u=17.0m/s.

Therefore, the initial velocity of the first ball must be 17.0m/s in order for both balls to hit the ground at the same time. This is because the first ball was in the air for a longer period of time, so it needed to have a slower initial velocity to cover the same distance as the second ball in less time.

In conclusion, as a scientist, I would use the relevant equations and variables to solve this problem and determine the initial velocity of the first ball.