2 objects 1 up and 1 down thrown

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An object is thrown downward at 9 m/s from a height of 40 m, while another is propelled upward from ground level at 14 m/s, with gravity at 9.8 m/s². To find the height where they meet, the distances traveled by both objects must be equalized using the kinematics formula. The calculations involve setting up the equation for each object's motion and solving for time. After determining the time, the distance can be calculated to find the height above the ground where the two objects pass each other. The final answer should be expressed in meters.
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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?
 
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christinono said:
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.
 
i guess i just can't 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?
 
Remember, the one being thrown down has a NEGATIVE velocity (negative direction).
 
ok gotcha i think, thanks for all the help
 
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
 
runner1738 said:
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
 
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