Three balls meeting at maximum height with vertical motion

[Joze]

Homework Statement

So I've been stuck on this exercise for a few hours now, maybe you can help out:

3 balls meet at the same height h_m.
Ball 1 is accelerated straight up into the air for 1s from height 0.
When the acceleration stops, ball 2 is launched straight up from height 0 with v₂ = 29.358 m/s.
A short time later ball 3 is dropped from height h_c=60m.
Each ball has a mass of 1g. The height h_m is the maximum height that ball 1 reaches on its flight path.
Friction can be neglected.
1) calculate h_m and the acceleration a of ball 1.
2) what is the time between the moment, that ball 2 is launched and the moment ball 3 is dropped from the ceiling h_c=60m?

Homework Equations

Kinematic equations for constant acceleration

The Attempt at a Solution

I defined:
t₀ as the time that the acceleration of ball 1 stops and ball 2 is launched with v₂ = 29.358 m/s
t₁ as the time that ball 3 is dropped from the ceiling = ?
t₂ as the time at which all balls are at height h_m = ?
h_m as the height at which all balls meet = ?
h₁ as the height at which the acceleration of ball 1 stops = ?
h_c as the height of the ceiling = 60m
v₁ as the velocity of ball 1 when the acceleration stops = ?
v₂ as the initial velocity of ball 2 = 29.358 m/s

Then I tried to come up with kinematic equations for all 3 balls:
- Ball 1:
eq1: h_m = h₁ + v₁*t₂ - 0.5*g*t₂²
eq1.1: v₁² = 2*g*(h_m-h₁)
- Ball 2:
eq2: h_m = v₂*t₂ - 0.5*g*t₂²
- Ball 3:
eq3: h_m = h_c - 0.5*g*(t₂ - t₁)²

I wrote the height of ball 1 as a function of time:
h(t) = -0.5*g*t² + v₁*t +h₁

The height at which the balls meet is the max point of the function above so I set the first derivative = 0 and solved for t:
h'(t) = -g*t + v₁ = 0
So t = v₁/g which should be t₂

I tried out plugging that into equation 1 and 2 and I also tried to set eq1 = eq2 to solve for something, but from that point I just went in circles and I just can't figure out how to proceed. Maybe I'm blind to something obvious because I've been trying for so long to solve this.

I would really appreciate any help or hints on how to go on from there.
Thanks in advance for your time.

 

[Ray Vickson]

Homework Statement

So I've been stuck on this exercise for a few hours now, maybe you can help out:

3 balls meet at the same height h_m.
Ball 1 is accelerated straight up into the air for 1s from height 0.
When the acceleration stops, ball 2 is launched straight up from height 0 with v₂ = 29.358 m/s.
A short time later ball 3 is dropped from height h_c=60m.
Each ball has a mass of 1g. The height h_m is the maximum height that ball 1 reaches on its flight path.
Friction can be neglected.
1) calculate h_m and the acceleration a of ball 1.
2) what is the time between the moment, that ball 2 is launched and the moment ball 3 is dropped from the ceiling h_c=60m?

Homework Equations

Kinematic equations for constant acceleration

The Attempt at a Solution

I defined:
t₀ as the time that the acceleration of ball 1 stops and ball 2 is launched with v₂ = 29.358 m/s
t₁ as the time that ball 3 is dropped from the ceiling = ?
t₂ as the time at which all balls are at height h_m = ?
h_m as the height at which all balls meet = ?
h₁ as the height at which the acceleration of ball 1 stops = ?
h_c as the height of the ceiling = 60m
v₁ as the velocity of ball 1 when the acceleration stops = ?
v₂ as the initial velocity of ball 2 = 29.358 m/s

Then I tried to come up with kinematic equations for all 3 balls:
- Ball 1:
eq1: h_m = h₁ + v₁*t₂ - 0.5*g*t₂²
eq1.1: v₁² = 2*g*(h_m-h₁)
- Ball 2:
eq2: h_m = v₂*t₂ - 0.5*g*t₂²
- Ball 3:
eq3: h_m = h_c - 0.5*g*(t₂ - t₁)²

I wrote the height of ball 1 as a function of time:
h(t) = -0.5*g*t² + v₁*t +h₁

The height at which the balls meet is the max point of the function above so I set the first derivative = 0 and solved for t:
h'(t) = -g*t + v₁ = 0
So t = v₁/g which should be t₂

I tried out plugging that into equation 1 and 2 and I also tried to set eq1 = eq2 to solve for something, but from that point I just went in circles and I just can't figure out how to proceed. Maybe I'm blind to something obvious because I've been trying for so long to solve this.

I would really appreciate any help or hints on how to go on from there.
Thanks in advance for your time.

Ball 1 accelerates at rate $a$ for the first second, so at time $t=1$ is at height $a/2$ and has upward speed $a$. (Easiest: let $a$ be dimensionless, so the initial acceleration is $a$ m/sec^2, and let time be dimensionless, so that time is $t$ sec.) Anyway, for $t > 1$ the altitude of Ball 1 is $h_1 = a/2 + a(t-1) - 1/2 g (t-1)^2$. Set the t-derivative to 0 to get $t_m,$ the maximum-height time. You can also get the maximum height $h_m$ by substituting $t=t_m$ into the equation for $h$. So far, everything will still depend on the unknown acceleration $a$.

At time $t > 1$ the height of Ball 2 is $h_2 = v_0 (t-1) - 1/2 g (t-1)^2$ (where $v_0 = 29.358$), and this must equal $h_m$ at $t = t_m$. That gives you an equation in $a$ which you can solve.

 

[Joze]

Thank you SO much!
Expressing the height and velocity of ball 1 with dependency on a was the missing piece.
Solved it now with your hints :)
Have a nice day!