Discussion
Where does he land?
*This question is included in 04. Gravitation, question #47
| (A) | 2.5 meters from the edge of the building, that is, on the pavement. |
| (B) | ... |
| (C) | ... |
| (D) | ... |
| (E) | ... |
| (F) | ... |
The solution is
Posted: 07/19/2013 18:49
Could u do a step by step of this one
Posted: 07/22/2013 16:02
Step by step of this one?
The explanation pretty much says it all. Since we are calculating the horizontal distance,
Δx = v1 * Δt + 1/2 a * (Δt)^2
We already calculated Δt in the previous problem as 1.2s. And since there is no horizontal force acting on the student, horizontal acceleration or a = 0. So, we are left with Δx = v1 * Δt = 5m/s * 1.2s = 6m.
The explanation pretty much says it all. Since we are calculating the horizontal distance,
Δx = v1 * Δt + 1/2 a * (Δt)^2
We already calculated Δt in the previous problem as 1.2s. And since there is no horizontal force acting on the student, horizontal acceleration or a = 0. So, we are left with Δx = v1 * Δt = 5m/s * 1.2s = 6m.