Two children are playing on a 178.2 m tall bridge. One child drops a rock (initial velocity zero) at t = 0 . The other waits 2.7 s and then throws a rock downward with an initial speed v0. If the two rocks hit the ground at the same time, what is v0 ?
for the first rock i used:
t(ground)=square root((2yo)/g)
... then I subtracted 2.7s from that number
I think you use Vo=(y-0)+.5gt^2/t
...I am not getting the correct answer. Any advice?
I agree that the dropped rock takes sqrt(2Yo/g) = 6.03 s to teach the ground. The thrown rock must take 6.03 - 2.7 = 3.33 s from the timne it is thrown.
178.2 = Vo*t - (g/2)t^2
Solve for Vo.
Vo = (178.2 + 4.9 t^2)/t
With parentheses in the right place, and t = 3.33 s, you should get the right answer.
does my answer need to be negative due to the rocks falling?
-37.19 m/s
To solve this problem, you can use the kinematic equations of motion in the vertical direction. Let's break down the steps to find the value of v0 accurately:
1. Calculate the time it takes for the first rock to hit the ground:
- Use the equation y = y0 + v0t - (1/2)gt^2, where
- y is the final displacement (in this case, 178.2 m),
- y0 is the initial displacement (in this case, 0 m),
- v0 is the initial velocity (which is 0 for the first rock),
- g is the acceleration due to gravity (approximately 9.8 m/s^2),
- t is the time it takes for the rock to hit the ground (which we want to find).
Solving the equation for t, we get:
178.2 = 0 + 0*t - (1/2)(9.8)(t^2)
Simplify the equation:
178.2 = -4.9t^2
Rearranging the equation:
4.9t^2 = -178.2
Divide both sides by 4.9:
t^2 = -178.2 / 4.9
Take the square root of both sides to find t:
t = √(-178.2 / 4.9) [Note: Since we are dealing with square roots of negative numbers, there is no real solution to this equation. Therefore, the first rock does not hit the ground.]
2. Now, let's proceed with the second rock:
- We know that the second rock is dropped with an initial velocity v0 after a delay of 2.7 seconds.
- Therefore, the time taken for the second rock to hit the ground will be 2.7 seconds less than the first rock (which didn't hit the ground).
3. Calculate the time it takes for the second rock to hit the ground:
- Use the equation of motion mentioned above, but this time consider the time as (t - 2.7) since the second rock is thrown after a delay of 2.7 seconds.
- Therefore, the equation becomes:
178.2 = 0 + v0(t - 2.7) - (1/2)(9.8)((t - 2.7)^2)
Simplify the equation:
178.2 = v0(t - 2.7) - 4.9(t^2 - 5.4t + 7.29)
Rearranging the equation:
178.2 = -4.9t^2 + (12.61 - 5.4v0)t + (7.29 - 2.7v0)
Now, we have a quadratic equation in terms of t. We can solve for t by equating the equation to zero and then solving for t using the quadratic formula.
4. Use the quadratic formula to solve for t:
- The quadratic formula is:
t = (-b ± √(b^2 - 4ac)) / (2a)
Comparing with our equation:
-4.9t^2 + (12.61 - 5.4v0)t + (7.29 - 2.7v0) = 0
Here, a = -4.9, b = (12.61 - 5.4v0), and c = (7.29 - 2.7v0).
Substitute these values into the quadratic formula and solve for t.
5. Once you find the value of t, substitute it back into the equation: 178.2 = 0 + v0(t - 2.7) - (1/2)(9.8)((t - 2.7)^2).
6. Solve this equation for v0 to find the velocity at which the second rock is thrown downward.
By following these steps, you should be able to accurately calculate the value of v0.