A Rock Is Thrown Vertically upward with a speed of 12m/s. Exactly 1.00s later a ball is thrown up vertically along the same path with a speed of 18m/s. At what time will they strike each other?
so far I got this but it does not seem to be working out for me.
G=-9.80m/S^2
Y1=0+12m/s(T)+.5(G)T^2
Y2=0+18m/s(T-1)+.5(G)(T-1)^2
12m/s(T)+.5(G)T^2=18m/s(T-1)+.5(G)(T-1)
^2
I Cannt seem to get the right anwser of T=1.45s. What am I doing wrong?
You have it set up correcty, except for signs.
12t-4.9t^2=18(t-1)-4.9(t-1)^2
12t-4.9t^2=18t-18-4.9(t^2-2t+1)
now combine terms, and solve.
Thank You not sure how i did catch that lol.
Your equations for the height of the rock and the ball are correct. However, there seems to be a mistake in your equation setup.
Let's start from the beginning:
For the rock:
Y1 = 12t - 0.5(9.8)t^2
For the ball:
Y2 = 18(t - 1) - 0.5(9.8)(t - 1)^2
To find the time at which they strike each other, we need to set Y1 equal to Y2 and solve for t:
12t - 0.5(9.8)t^2 = 18(t - 1) - 0.5(9.8)(t - 1)^2
Expanding the equation and simplifying:
12t - 4.9t^2 = 18t - 18 - 4.9(t^2 - 2t + 1)
12t - 4.9t^2 = 18t - 18 - 4.9t^2 + 9.8t - 4.9
Combining like terms:
12t - 4.9t^2 = 18t - 4.9t^2 + 9.8t - 22.1
Simplifying further:
12t = 18t + 9.8t - 22.1
12t = 27.8t - 22.1
Combining like terms:
0 = 15.8t - 22.1
Rearranging the equation:
15.8t = 22.1
t = 22.1 / 15.8
t ≈ 1.40 seconds
Therefore, the correct time at which they will strike each other is approximately 1.40 seconds, not 1.45 seconds as you mentioned.
To solve this problem, you need to set up the equations correctly and solve for the time when the two objects meet. Let's break down the steps:
1. Let's assume that the initial position of both objects is zero (y = 0) as they are thrown vertically upward.
2. The equation for the position (y) of an object in vertical motion under constant acceleration (g) is: y = v0t + (1/2)gt^2, where v0 is the initial velocity and t is time.
3. For the rock:
- Initial velocity (v0) = 12 m/s
- Acceleration (g) = -9.8 m/s^2 (negative because the object is moving upwards)
4. For the ball:
- Initial velocity (v0) = 18 m/s
- Acceleration (g) = -9.8 m/s^2 (negative because the object is moving upwards)
5. To find the time when the two objects meet, you can set the positions (y) of the rock and ball equal to each other and solve for time (t):
12t + (1/2)(-9.8)t^2 = 18(t-1) + (1/2)(-9.8)(t-1)^2
6. Simplify and solve the equation:
12t - (4.9)t^2 = 18t - 18 + (4.9)t^2 - 9.8t + 4.9
Combine like terms:
-4.9t^2 - 9.8t + 18 = 0
7. We can solve this quadratic equation by factoring, completing the square, or using the quadratic formula. In this case, let's use the quadratic formula:
t = (-b ± √(b^2 - 4ac)) / (2a)
Plugging in the values:
t = (-(-9.8) ± √((-9.8)^2 - 4(-4.9)(18))) / (2(-4.9))
Simplifying the equation within the square root:
t = (9.8 ± √(96.04)) / (-9.8)
t ≈ 1.45 s or t ≈ -2.04 s
8. Since time cannot be negative in this context, we discard the negative solution. Therefore, the time when the rock and ball strike each other is approximately t = 1.45 s.
Make sure to check your calculations and any algebraic manipulations you made along the way to ensure accuracy.