- #1
blacksilver5
- 2
- 0
1. A person drops a ball from the top of a building; 1 second later, a second ball is thrown straight up from the bottom of the same building. The ball that is thrown upwards has just the necessary speed to just make it to the top of the building.
As the balls pass each other, they have the same speed. Half a second after they pass, the ball that was dropped hits the ground. What was the height of the building?
2. I believe that this is a multi-step problem, but it uses either
Displacement=S * t + .5 * a * t^2 where S-initial speed, t-delta time, a-acceleration
or S[f]^2 - S^2 = 2 * a * D where S[f]-final speed, D-displacement
3. I'm not sure, but i drew a pic and know that if B1 is the one being dropped, B2 is thrown upward at t=1, Speed of B1 and B2 are eqaul at time x, at time x+.5 B1 hits the ground, the distance traveled by B1 equals the height of the building
As the balls pass each other, they have the same speed. Half a second after they pass, the ball that was dropped hits the ground. What was the height of the building?
2. I believe that this is a multi-step problem, but it uses either
Displacement=S * t + .5 * a * t^2 where S-initial speed, t-delta time, a-acceleration
or S[f]^2 - S^2 = 2 * a * D where S[f]-final speed, D-displacement
3. I'm not sure, but i drew a pic and know that if B1 is the one being dropped, B2 is thrown upward at t=1, Speed of B1 and B2 are eqaul at time x, at time x+.5 B1 hits the ground, the distance traveled by B1 equals the height of the building
