Hello, I had a question for physics, and I tried it out, however, I got the wrong answer.
The Question:If a rock takes 0.750s to hit the ground after being thrown down from a height of 4.80m, determine the rock's initial velocity.
My Work : The formula- d=vi*t+1/2*a*t^2
The isolated formula-(2*d)/(t*a)=vi
So it would be (2*4.80m)/(0.750s*9.81m/s^2)=vi
1.30m/s=vi
The correct answer is 2.72m/s[down]/
You you please show me where I went wrong? Thank you!
Heres your answer by the Verified user.
your formula is incorrect the right formula is as defined
[Formula used]
d=(ViT)+(1/2at^2)
vi=(0.75/4.8)+(1/2*9.8*0.75^2)
Vi = (2.9)Answer
Heres your answer by the Verified user.
your formula is incorrect the right formula is as defined
[Formula used]
d=(ViT)+(1/2at^2)
vi=(0.75/4.8)+(1/2*9.8*0.75^2)
Vi = (2.9)Answer
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To determine the rock's initial velocity, you applied the correct formula d = vi * t + 1/2 * a * t^2, where d is the height of the rock, vi is the initial velocity, t is the time of fall, and a is the acceleration due to gravity.
However, there seems to be a mistake in your calculation. The isolated formula should be vi = (2 * d - a * t^2) / (2 * t), not (2 * d) / (t * a).
Let's plug in the values given:
d = 4.80 m
t = 0.750 s
a = 9.81 m/s^2
Using the corrected formula:
vi = (2 * 4.80 m - 9.81 m/s^2 * (0.750 s)^2) / (2 * 0.750 s)
Now, we can solve this equation:
vi = (9.60 m - 6.953 m) / 1.50 s
vi = 2.647 m / 1.50 s
vi ≈ 1.765 m/s
So, it appears that you made a calculation error in your work. The correct initial velocity is approximately 1.765 m/s. It seems the answer you provided, 2.72 m/s (down), might be incorrect.
Hello! It seems like you've made a small error in your calculation. Let's go through the problem together and find where the mistake might have occurred.
The formula you used, d = vi * t + (1/2) * a * t^2, is correct. However, you made an error when isolating the formula to solve for vi.
Instead of using (2*d)/(t*a) = vi, you should have used (2*d)/(t^2) = a + vi. This is because you want to isolate vi, not a.
So, let's correct the calculation:
Given:
d = 4.80 m
t = 0.750 s
a = 9.81 m/s^2 (acceleration due to gravity)
Using the corrected formula, we have:
(2 * d) / (t^2) = a + vi
Substituting the given values:
(2 * 4.80 m) / (0.750 s)^2 = 9.81 m/s^2 + vi
Calculating the left side:
(2 * 4.80 m) / (0.750 s)^2 = 17.0667 m/s^2
Simplifying the equation:
17.0667 m/s^2 = 9.81 m/s^2 + vi
Now, solving for vi:
vi = 17.0667 m/s^2 - 9.81 m/s^2
vi = 7.2567 m/s
So, the correct initial velocity (vi) of the rock is approximately 7.26 m/s, not 1.30 m/s as you calculated.
I hope this clarifies the issue and helps you find where you went wrong. If you have any further questions, feel free to ask!