Vertical Motion 1

Leung

2015-04-29 5:55 pm

The attached link show a v-t graph of a stone being projected vertically upwards.

Can anyone vertify whether my understanding is correct ? Tks

1. The upward direction is taken as postive (T)
2. The stone reaches the highest point at P (F - should be Q)
3. Slope of graph is zero at Q (F - there is still gravity acceleration)
4. Velocity changes in direction at Q (T)
5. Stone first returns to its inital position at R (?? - this is a v-t graph, how to
(estimate the displacement)
6. Maximum height reached is 5m (??)

Tks

https://www.sendspace.com/file/0s7v1a

回答 (1)

天同

2015-04-29 6:29 pm

✔ 最佳答案

1. Correct.

2. Not correct. The highest point occurs at Q.

3. Not correct.

4. Correct.

5. Correct.
This is based on two reasons:
(i) The stone takes 1 s to go up from the projecting position to the highest point (at Q), and then a further 1 s to fall down from the highest point. Because of the symmetry of the two paths (going up and falling down paths). Point R is the original projecting point.

(ii) The displacements are given by the areas under the two triangles, one above the time-axis and one below the time-axis. Because the areas of the two triangles are the same, this shows that the upward displacement equals to the downward displacement. Hence, point R is the initial point.

6. You can't tell, because the speed of projection is not given.

2015-04-29 10:39:30 補充：
sorry, we can calculate the displacement in (6):
Given v = 0 m/s, a = -g(= -10 m/s^2), t = 1 s , s =?
Using s = vt - (1/2)at^2
thus, s = -(1/2).(-10).(1^2) m = 5 m

2015-04-29 10:40:18 補充：
Therefore, point (6) is correct.

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