Ball Dropping vs Throwing (1d Kinematics)

[Kildars]

A ball is thrown upward from the ground with an initial speed of 30 m/s; at the same instant, a ball is dropped from a building 20 m high. After how long will the balls be at the same height?

Again, some guidance on where to start would help me crack this one better.

 

[BishopUser]

$$y=y_0+v_0t+\frac {1} {2}at^2$$
Use this equation for both balls using the same zero level. Their Y levels will be the same so you can put both equations together and solve for t.

 

[Kildars]

$$y=y_0+v_0t+\frac {1} {2}at^2$$
Use this equation for both balls using the same zero level. Their Y levels will be the same so you can put both equations together and solve for t.

Is y 20? and y0 = 0? I know that Vo is 0.

 

[BishopUser]

y = the position at which the 2 balls meet
y0 is the initial position. If you set the 0 level at the ground then the ground ball y0 = 0 and the building ball y0 = 20. The building ball will have v0 = 0, the ground one will have v0=30m/s. Make 2 different equations with that information, combine then, solve for T

 

[Kildars]

Building Ball y = 20 + 0t + 1/2(-9.81)t^2
Ground Ball y = 0 + 30t + 1/2(-9.81)t^2

Do I combine them by doing y = 20 + 30t + 1/2(-9.8)t^2

 

[BishopUser]

both equations are equal to the same y, so you combine them by setting them equal to one another

20 + 0t + 1/2(-9.81)t^2 = 0 + 30t + 1/2(-9.81)t^2
20 = 30t
t = 2/3 seconds

 

[Kildars]

both equations are equal to the same y, so you combine them by setting them equal to one another

20 + 0t + 1/2(-9.81)t^2 = 0 + 30t + 1/2(-9.81)t^2
20 = 30t
t = 2/3 seconds

Thank you, you learn something everyday :)