Thread: Two rocks passing each other

Originally Posted by Elite_Guard89

A rock is thrown upward from ground level with an initial velocity of 15 m/s while at the same instant another rock is thrown downward from a bridge that is 20m above the ground. The second rock has an initial velocity of 8 m/s downward. Ignore air resistance. Find the time when they pass by each other. How high above the ground are the rocks when they pass each other? How fast are the rocks moving when they pass each other?

For time, I ended up getting .204s but then again I was making assumptions when I did that calculation. Can anybody point me in the right direction?

Up rock (take upwards direction as positive): u = 15 m/s, a = -9.8 m/s^2, s = y.

Down rock (take downwards direction as positive): u = 8 m/s, a = 9.8 m/s^2, s = x.

When rocks pass each other, t is the same for each and x + y = 20m:

y = 15t - 4.9 t^2 .... (1)

x = 8t + 4.9 t^2 .... (2)

x + y = 20 .... (3).

Substitute (3) into (1):

20 - x = 15t - 4.9 t^2 .... (1')

x = 8t + 4.9 t^2 .... (2')

Solve (1') and (2') simultaneously for x and t:

(1') + (2'): 20 = 23 t => t = 20/23.

Substitute t = 20/23 and solve for x and hence y.


Up rock (take upwards direction as positive): u = 15 m/s, a = -9.8 m/s^2, s = y = ....., v = ?

Down rock (take downwards direction as positive): u = 8 m/s, a = 9.8 m/s^2, s = x, ....., v = ?

Solve for final velocity v of each rock using an appropriate formula.

Plugging in (20/23) x came out to be 3.25m. If x+y=20 then y should be 16.75 correct? Meaning that they cross at 16.75m above the ground?

Originally Posted by Elite_Guard89

Plugging in (20/23) x came out to be 3.25m. If x+y=20 then y should be 16.75 correct? Meaning that they cross at 16.75m above the ground?

How can you substitute t = 20/3 into x = 8t + 4.9 t^2 and get x = 3.25??

I guess I put it in the calculator wrong lmao. Nevermind!