I have a few questions that I need help with and that need to be checked please.
t takes 6 seconds for a stone to fall to the bottom of a mine shaft. How deep is the shaft? (I know I have to use this formual d=vinitial+1/2gt^2, I plug in these numbers d=0+1/2(9.8m/s^2)(6^2 s= 176.4 m, is this right?)
If a projectile is fired straight up at a speed of 10 m/s, the total time to return to its starting position is about (I know I have to use this formula, hfinal=hinitial+vinitial *time+ 1/2gt^2, these are the numbers I plugged in hfinal=0+0*10m/s+1/2(9.8m/s^2(10m/s)^2=490 s, when I suppose to get 2 seconds for my answer.
Ten seconds after starting from rest, an object falling freely downward will have a speed of about ? (I know I have to use this formual vfinal=gt, here is what I plugged in vfinal=(9.8m/s^2)(10 s)=98 m, when I am suppose to get 1000 m/s
A man leans over the edge of a cliff and throws a rock upward at 4.9m/s. How far below the level from which it was thrown is the rock 2 seconds later? ( would the formula I use be d=vt?
yes on the stone falling.
No on the second.
Hfinal=hinitial+ vinitial*t -4.9t^2
0=0+10t-4.9t^2 solve for t
The third: vfinal=gt= 9.8m/s^2*10sec=98m/s. It wont be 1000m/s
The final:
Hfinal=Hinitial+Vi*t - 4.9t2
Let Hintialal be 0
Let's go through each question and check the calculations:
1) To find the depth of the shaft, you correctly used the formula d = v_initial * t + (1/2) * g * t^2. You plugged in the values as d = 0 + (1/2) * (9.8 m/s^2) * (6^2 s) = 0 + 1/2 * 9.8 * 36 = 0 + 176.4 = 176.4 m. So, your answer of 176.4 m is correct.
2) For the projectile fired straight up, to find the total time to return to its starting position, you used the formula h_final = h_initial + v_initial * time + (1/2) * g * t^2. However, you made a slight calculation mistake. Plugging in the values correctly, we have h_final = 0 + 0 * t + (1/2) * 9.8 * (10)^2. Evaluating this, h_final = (1/2) * 9.8 * 100 = 490 m. So, your answer of 490 m is correct.
3) For an object falling freely downward, you used the formula v_final = g * t. You plugged in the values correctly, giving v_final = 9.8 * 10 = 98 m/s. So, your answer of 98 m/s is correct.
4) For the rock thrown upward from the cliff, you need to consider the upward trajectory and the downward motion separately. To find the distance below the level from which it was thrown after 2 seconds, you need to calculate the displacement during the upward motion and then during the downward motion.
During the upward motion, you can use the formula v_final = v_initial + g * t. Plugging in the values, we have v_final = 0 + (-9.8 m/s^2) * 2 = -19.6 m/s.
After 2 seconds, the rock reaches its maximum height and starts falling downward. To find the displacement during the downward motion, you can use the formula d = v_initial * t + (1/2) * g * t^2. Plugging in the values, we get d = 4.9 m/s * 2 s + (1/2) * 9.8 m/s^2 * (2 s)^2 = 9.8 m + 19.6 m = 29.4 m.
Therefore, the rock is approximately 29.4 meters below the level from which it was thrown after 2 seconds.