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runner1738
Feb4-05, 08:16 PM
An object is thrown downward with an initial speed of 9 m/s from a height of 40 m above the ground. At the same instant, a second object is propelled vertically from ground level with a speed of 14 m.s. The acceleration of gravity is 9.8 m/s^2. At what height above the ground will the two objects pass each other? Abswer in units of m?

christinono
Feb4-05, 08:20 PM
Use the following kinematics formula and equal the distances they travel. Then, solve for time.

d = V_it+ \frac{1}{2}at^2
Same idea as this one. Equal the distances.

runner1738
Feb4-05, 10:32 PM
i guess i just cant do math in general casue i got 9t + 1/2(9.8)t^2=14t+1/2(-9.8)t^2
9.8t^2-5t=0 , that i got .5102040816 * what though?

christinono
Feb4-05, 10:34 PM
Remember, the one being thrown down has a NEGATIVE velocity (negative direction).

runner1738
Feb4-05, 11:40 PM
ok gotcha i think, thanks for all the help

runner1738
Feb4-05, 11:51 PM
ok this time around i got 2.346938976 from -9t+1/2(9.8)t^2=14t+1/2(-9.8)t^2, but the answer is supposed to be in m? and i have time

christinono
Feb5-05, 01:47 PM
ok this time around i got 2.346938976 from -9t+1/2(9.8)t^2=14t+1/2(-9.8)t^2, but the answer is supposed to be in m? and i have time
From the time, you can find the distance. Just use the formula:

d=V_it + \frac{1}{2}at^2