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]/
Could you please show me where I went wrong? Thank you!
height = 1/2 g t^2 + Vo t + Ho
0 = -4.90 * .750^2 + Vo * .750 + 4.80
Hello! It seems like there was a mistake in your calculations. Your approach is correct, but there was an error in the equation you used to isolate the initial velocity.
Let's go through the problem step by step:
1. Start with the equation of motion: d = vi * t + (1/2) * a * t^2
2. Identify the values given in the problem:
- Initial height (d) = 4.80 m
- Time (t) = 0.750 s
- Acceleration due to gravity (a) = -9.81 m/s^2 (negative sign because it is downward)
3. Rearrange the equation to solve for the initial velocity (vi):
d - (1/2) * a * t^2 = vi * t
vi = (d - (1/2) * a * t^2) / t
4. Substitute the values into the equation:
vi = (4.80 m - (0.5 * (-9.81 m/s^2) * (0.750 s)^2)) / 0.750 s
5. Simplify the equation:
vi = (4.80 m + 0.5 * 9.81 m/s^2 * (0.750 s)^2) / 0.750 s
vi = (4.80 m + 0.5 * 9.81 m/s^2 * 0.5625 s^2) / 0.750 s
vi = (4.80 m + 2.69921875 m) / 0.750 s
vi = 7.49921875 m / 0.750 s
vi ≈ 9.99895833 m/s
It seems like you missed the negative sign for acceleration and made a calculation mistake while adding the terms. The correct answer for the initial velocity is approximately 9.999 m/s (upward).
I hope this helps! Let me know if you have any further questions.