Question

How long does it take a ball to reach the ground 7.0 m below if thrown straight up with initial speed of 2.00m/s?

i got

time=?
g=9.8m/s^2
iv=2.00m/s
vf=7.0m

should i use ivt plus 1/2at^2

Answer 1

7 meters below ground? Is it thrown up from the ground at 0 meters?

Answer 2

It is thrown from an elevated position 7ft and lands at a lower point 0ft, like throwing it upward from a roof and it landing in the alley below?

Answer 3

Hm, maybe we have to do factoring

Answer 4

2 m/s / 9.8 m/s^2 = .204s

D = vit + 1/2 at^2+Hi (solving for max height above ground)
D = 2 m/s * .204s + (-4.9 m/s^2) * (.204s)^2
D = .408 m + -.2039m + 7m = 7.204m

D = vit + 1/2at^2 (since at the top of the parabola, v = 0, vit can be cancelled)
-7.204m = -4.9 m/s^2 * t^2
1.47s^2 = t^2
t = 1.2124s

Answer 5

I'm using these formulas:
y = Voy * t - 1/2 *g * t^2
quadratic formula , where t = Voy/ g = sq rt [V^2oy / g^2 - 2y / g]
y = vertical distance
Voy = initial velocity
g = 9.81 m/s^2

t^2 gives two values for time: t = 1.41 sec, for one value and -1.008 sec.
Obviously time can't be negative so the time of flight is 1.41 seconds
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Answer 6

.-. work?

Answer 7

The answer I gave for 1.41 seconds, is from the time it takes launching 2 m/s at sin90 which takes 0.204 seconds to reach peak and coincidentally, it reaches a height of 0.204 meters.
Now the ball falls back down for 0.204m + 7 meters = 7.204m

To freefall for 7.204m takes 1.21 seconds. But you have to add the 0.204 s
when it reached a peak before it started trajectory down
So 1.21s + 0.204 s = 1.41 second total time

Answer 8

Ah, I see where I went wrong XD thanks, I forgot to add the previous time

Answer 9

i think so if the distance is the total distance travelled then you can use speed = distance/time