Ball Drop Problem: Find Initial Velocity

[ArbazAlam]

A ball is thrown straight up from the edge of the roof of a building. A second ball is dropped from the roof 2.00 s later. Air resistance may be ignored.

(a) If the height of the building is 60 m, what must be the initial speed of the first ball if both are to hit the ground at the same time?

I've done a number of things that have given me very strange(improbable) numbers. Now what I've attempted to do is set up an equation for each of the balls.

y₁(t) = -4.9(t+2)²
y₂(t) = v_ot - 4.9t²

I'm not sure where to go from here. I tried setting y₁ to 60 and solving to find the time and then plugging it into y₂ but that turned out to be incorrect.

 

[PhanthomJay]

You've got your times reversed and you must watch your plus and minus signs. The second ball is thrown later. If it takes the second ball , which you are calling 1, 't' seconds to reach the ground, the first ball has t + 2 seconds to reach the ground. Also, what is the direction of the displacement?

 

[baba89]

I would approach this problem by using the equations of motion to solve for the initial velocity of the first ball. The equations of motion are:

y(t) = y0 + v0t + 1/2at^2
v(t) = v0 + at

where y(t) is the position at time t, y0 is the initial position, v0 is the initial velocity, a is the acceleration, and t is time.

Using these equations, we can set up a system of equations for the two balls:

Ball 1 (thrown up):

y(t) = 60 m (since it must reach the ground at the same time as ball 2)
v(t) = v0 - 9.8 m/s^2 (since it is slowing down due to gravity)

Ball 2 (dropped):

y(t) = 0 m (since it starts at the top of the building and falls to the ground)
v(t) = -9.8 m/s^2 (since it is accelerating due to gravity)

Solving for t in both equations:

t = (60 - y0)/v0
t = y0/9.8

Since both balls must hit the ground at the same time, we can set these two equations equal to each other and solve for v0:

(60 - y0)/v0 = y0/9.8
v0 = 9.8(60 - y0)/y0

Substituting y0 = 60 m (since the building is 60 m tall):

v0 = 9.8(60 - 60)/60 = 0 m/s

This means that the initial velocity of the first ball must be 0 m/s in order for both balls to hit the ground at the same time. This makes sense since the first ball is thrown straight up and will reach a maximum height before falling back to the ground, while the second ball is simply dropped from the top of the building. Therefore, their initial velocities will be different and the first ball will take longer to reach the ground.

In conclusion, the initial velocity of the first ball must be 0 m/s in order for both balls to hit the ground at the same time.