A rock is dropped from a height of 100m.
a) how long does it take to fall the first 50m?
b)How long does it take to fall the last 50m?
Which equation would I use to solve this problems with?
For a) use s=(1/)g*t^2 where s=50m and g=9.8m/s^2 and solve for t.
For b)use s=(1/)g*t^2 where s=1000m and g=9.8m/s^2, solve for t and subtract the result of part a).
You use the same formula, but you look at the total time minus some partial time. Does that make sense?
A stone propelled from a catapult with a speed of 50m-s attains a height of 100m calcutate
Yes, that makes sense! To solve both parts of the problem, you can use the equation for motion under constant acceleration, which is known as the kinematic equation for displacement:
s = (1/2) * g * t^2
where:
- s is the displacement (distance traveled)
- g is the acceleration due to gravity (approximately 9.8 m/s^2 on Earth)
- t is the time taken to fall
Now let's apply this equation to each part of the problem:
a) For the first part, you want to find how long it takes for the rock to fall the first 50 meters. So, you have s = 50m. Rearrange the equation to solve for t:
s = (1/2) * g * t^2
50 = (1/2) * 9.8 * t^2
Dividing both sides by (1/2) * 9.8:
t^2 = 50 / ((1/2) * 9.8)
t^2 = 10.2041
Taking the square root of both sides:
t ≈ √(10.2041)
t ≈ 3.19 seconds
Therefore, it takes approximately 3.19 seconds for the rock to fall the first 50 meters.
b) For the second part, you want to find the time it takes for the rock to fall the remaining 50 meters after it has already fallen 50 meters. So, you have s = 50m (since you're looking at the remaining distance). Again, rearrange the equation to solve for t:
s = (1/2) * g * t^2
50 = (1/2) * 9.8 * t^2
Dividing both sides by (1/2) * 9.8:
t^2 = 50 / ((1/2) * 9.8)
t^2 = 10.2041
Taking the square root of both sides:
t ≈ √(10.2041)
t ≈ 3.19 seconds
As you can see, the result is the same as in part a), which means it takes the same amount of time for the rock to fall the second 50 meters.
However, to determine the time it takes for the last 50 meters specifically, you need to subtract the time it takes for the first 50 meters. So, subtract the result from part a) from the total time:
Total time = 3.19 seconds (calculated in both parts)
Last 50 meters time = Total time - Part a) time
Last 50 meters time = 3.19s - 3.19s
Last 50 meters time = 0 seconds
This means that it takes 0 seconds for the rock to fall the last 50 meters because it already fell during the time calculated in part a).
So, to summarize:
a) It takes approximately 3.19 seconds for the rock to fall the first 50 meters.
b) It takes 0 seconds for the rock to fall the last 50 meters.
Yes, that makes sense. To solve for the time it takes to fall the first 50m, you can use the equation s = (1/2)gt^2, where s is the distance fallen (50m), g is the acceleration due to gravity (9.8m/s^2), and t is the time taken.
Here's how to solve part a:
1. Plug in the given values into the equation: 50 = (1/2)(9.8)t^2
2. Simplify the equation: 100 = 9.8t^2
3. Divide both sides of the equation by 9.8: t^2 = 100/9.8
4. Take the square root of both sides to solve for t: t = √(100/9.8)
For part b, you want to find the time it takes to fall the last 50m. You can use the same equation, but instead of subtracting the result of part a from the total time, you can subtract the time it took to fall the first 50m.
Here's how to solve part b:
1. Plug in the given values into the equation: 50 = (1/2)(9.8)t^2
2. Simplify the equation: 100 = 9.8t^2
3. Divide both sides of the equation by 9.8: t^2 = 100/9.8
4. Take the square root of both sides to solve for t: t = √(100/9.8)
5. Subtract the result of part a from this value: t_b = √(100/9.8) - √(100/9.8 from part a)
So the equation used for both parts is s = (1/2)gt^2, where s is the distance fallen, g is the acceleration due to gravity, and t is the time taken.