Where is it at this time?
Plan: Remember that gravity is acting against the ball, so the acceleration a = -g = - (constant of gravity near Earth’s surface) = - 9.8 m/s2. Due to gravity, at some point the ball is going to stop (v=0 at the maximum height that the ball reaches) and start to fall back down. We will use +y direction as the up direction.
2b.Next, from the same position, you throw an identical ball (Ball 2) straight down with the same launch speed as in the previous question. How fast will the ball be traveling 2 seconds later and where is it at this time?
The reason for this is that ball 1 reaches its maximum height, then falls back down. When it crosses the point where the platform is, it has the same speed it initially had when thrown up, only now it is going downwards with the same speed.
Proof of this: Take ball 1, and find the final speed it would have if the final y value is the same as the initial y value (so that the ball is released from the platform, goes up, and comes right back to the platform). Use the equation vf2 = v02 -2g(yf-y0); the initial and final y values are the same, so their difference is 0 and we have vf2 = v02 -2g(0)= v02 in other words, vf and v0 have the same magnitudes, but different directions: vf = -v0
3.The ceiling of a classroom is 3.75 m above the floor. A student tosses an apple vertically upward, releasing it 0.5 m above the floor. What is the maximum initial speed that can be given to the apple if it is not to touch the