a mass is thrown vertically upwards with
an initial velocity of 30m/s. a second mass is
dropped 0.5s later from directly above a height of
60m from the first mass. when do they meet and
at what height?
You want to find where
30t - 4.9t^2 = 60 - 4.9(t-.5)^2
t = 2.34
now plug that in to find either height.
Assuming you meant 60m above where the first mass started. Things are different if the second mass is dropped when the first mass is 60m below it. But I'm sure you can make the adjustments needed. If not, c'mon back and show whatcha got.
To find when and at what height the two masses meet, we can use the equations of motion. Let's find the time at which they meet first:
For the mass thrown vertically upwards:
Initial velocity (u) = 30 m/s (upwards)
Acceleration (a) = -9.8 m/s² (downwards, due to gravity)
The displacement of the first mass (S₁) can be found using the equation: S₁ = ut + (1/2)at²
For the mass that is dropped:
Initial height (H) = 60 m
The displacement of the second mass (S₂) can be found using the equation: S₂ = H + (1/2)gt² (where g is acceleration due to gravity)
The two masses will meet when their displacements are equal. So, we have the equation: S₁ = S₂
Using the above equations, we can solve for the time (t) at which they meet:
S₁ = ut + (1/2)at²
S₂ = H + (1/2)gt²
So, 30t + (1/2)(-9.8)t² = 60 + (1/2)(-9.8)t²
Simplifying the equation: 15t - 4.9t² = 60
Now, let's solve this equation to find the value of t.