Negative Acceleration?

Question

A stone is thrown straight upwards with an initial speed of 47.5 m/s. (a) How long is the stone in the air before it hits the ground? (b) How high does it rise? It is assumed that the stone heads upwards from ground level.

I got:

(a) 9.7 sec

(b) 115.2 m

Am I correct? I hope I used the right number for acceleration (-9.8m/s^2). Thanks.

Answer 1

I've already answered the g > 0 or g < 0 question. Just be consistent with all the vectors: s, u, v, and g or a.

Your tossed stone travels two paths: up and down.

Up, we have s = 1/2 gt^2 = 1/2 v^2/g (from the previous question you asked); so that s ~ 1/2 25 X 10^2/10 = 12.5 X 10 m, which is approximately what you got for the height the stone reaches above ground.

Also up. Time to reach v = 0 = u - gt; so that t = u/g ~ 4.75 sec where u = 47.5 m/sec initial velocity. [NB: The initial velocity u > 0 to indicate upward; so g < 0 is true because gravity's acceleration is downward. Just ask yourself, is gravity speeding or slowing the stone? Clearly the stone is slowing because its velocity at the apex of the toss is v = 0.]

Now down. Because the stone falls under the same g it experience going up, the time to fall will be the same as the time to rise to its apex at about s = 125 meters. Therefore, the total time of travel (up and down) is T = 2t ~ 2*4.75 = 9.5 sec, which is also approximately what you got.

Nice job.

Answer 2

For this problem, you should use g = 9.81 m/sec² since the precision of the initial speed is to 3 sig figs.

t(up) = Vi/g
...= 47.5/9.81
...= 4.84 sec

t = t(up + t(down)
...= 4.84 * 2
...= 9.68 sec

H = 1/2gt(up)²
... = 1/2 * 9.81* 4.84² = 115 m

All answers to 3 sig figs