A ball is thrown vertically downward with a velocity of 12m/s How far has the ball fallen 1 s later?. 2s later?
h=vₒt+gt²/2,
v=12
t=1
g=9.9
Well, well, well, it looks like the ball is in a bit of a hurry to hit the ground! Let me calculate the distances for you.
After 1 second, the ball would have covered a distance of approximately 12 meters. Why? Because it likes to make a dramatic entrance, so it keeps up its speed!
Now, after 2 seconds, the ball is getting closer to the ground. It would have fallen a whopping distance of approximately 48 meters. Talk about gravity bringing it down to earth fast!
Just remember, my calculations might not be perfect, but they're definitely funny. So, take them with a sprinkle of giggles!
To determine the distance the ball has fallen after a certain amount of time, we can use the formula for vertical distance traveled:
d = v * t + (1/2) * g * t^2
Where:
- d is the distance traveled
- v is the initial velocity
- t is the time elapsed
- g is the acceleration due to gravity (approximately 9.8 m/s^2)
Let's calculate the distance the ball has fallen after 1 second:
d = (12 m/s) * (1 s) + (1/2) * (9.8 m/s^2) * (1 s)^2
d = 12 m + (1/2) * 9.8 m/s^2
d = 12 m + 4.9 m
d = 16.9 m
Therefore, the ball has fallen approximately 16.9 meters after 1 second.
Now, let's calculate the distance the ball has fallen after 2 seconds:
d = (12 m/s) * (2 s) + (1/2) * (9.8 m/s^2) * (2 s)^2
d = 24 m + (1/2) * 9.8 m/s^2 * 4 s^2
d = 24 m + 19.6 m
d = 43.6 m
Therefore, the ball has fallen approximately 43.6 meters after 2 seconds.
To find the distance the ball has fallen at different time intervals, we can use the equation of motion for free-falling objects:
s = ut + (1/2)gt^2
where:
s is the distance fallen
u is the initial velocity
t is the time elapsed
g is the acceleration due to gravity (approximately 9.8 m/s^2)
Let's calculate the distance the ball has fallen 1 second later:
Given: initial velocity (u) = 12 m/s, time (t) = 1 s
Using the formula:
s = ut + (1/2)gt^2
Substituting the values:
s = (12 m/s)(1 s) + (1/2)(9.8 m/s^2)(1 s)^2
s = 12 m + (1/2)(9.8 m/s^2)(1 s)^2
s = 12 m + (1/2)(9.8 m)(1)
s = 12 m + 0.5(9.8 m)
s = 12 m + 4.9 m
s = 16.9 m
Therefore, the ball has fallen a distance of 16.9 meters after 1 second.
Now let's calculate the distance the ball has fallen 2 seconds later:
Given: initial velocity (u) = 12 m/s, time (t) = 2 s
Using the formula:
s = ut + (1/2)gt^2
Substituting the values:
s = (12 m/s)(2 s) + (1/2)(9.8 m/s^2)(2 s)^2
s = 24 m + (1/2)(9.8 m/s^2)(4 s)^2
s = 24 m + (1/2)(9.8 m)(16)
s = 24 m + 78.4 m
s = 102.4 m
Therefore, the ball has fallen a distance of 102.4 meters after 2 seconds.
d=vo*time+1/2 g t^2
vo=-12m/s
g=-9.8m/s^2
find d (1sec)
I am not certain if you want the total distance, or distance in the ,2 sec. If that,
d=d(1.2)-d(1)