Solving for Time and Height of Sphere Meeting | 1 D Motion Problem

[medinat16]

Homework Statement

A sphere is thrown upwards with an initial velocity of 30m/s. A second sphere is dropped from directly above, a height of 60m from from the first sphere, 0.5 seconds later.When do the spheres meet and how high is the point where they meet?Please explain how you got your answer. My teacher got an answer of 1.68sec as the time and 38m as where they meet.

Homework Equations

xf=xi+vit+1/2at^2
vf^2-vi^2=2Δx(a)

The Attempt at a Solution

vi=30m/s
Δx=60m
t=0.5 sec
a=-9.8m/s^2

 

[hjelmgart]

How do you interpret the "0.5 seconds later". So is the second sphere 60 m from the first sphere, when it is dropped, or is it 60 m from the sphere after the 0.5 seconds?Nonetheless, I do not get the same result as your teacher. I put

xf=xi+vit+1/2at^2

For sphere 1 equal to that of sphere 2, since we want to know the time, where xf_1 = xf_2

I calculated a new initial velocity and a new position for sphere 1 after 0.5 sec with:

vf=vi+a*t = 25.1 m/s

Then the position xf = (vi+vf)*t/2 = 13.775 m

xf_1 = xf_2

xi+vit+1/2at^2 = xi+vit+1/2at^2

13.775m+25.1m/s t+1/2(-9.8m/s^2)t^2 = 60m+0t+1/2(-9.8m/s^2)t^2

solving for t gives 1.84s

I don't see, how this is wrong, perhaps someone else gets the same?

 

[burnst14]

I agree with your calculations and I obtain 43.37m as the point of contact using that assumption.
If we calculate assuming the second sphere is 60m away at 0.5 seconds, then:

0m+25.1m/s t+1/2(-9.8m/s²)t² = 60m+0m/s t+1/2(-9.8m/s²)t²

t = 2.39 s
and x_f = 32.01m

Still not your teacher's value.