Projectile Motion of thrown rock

[Momentum09]

1. You are standing on the edge of a ravine that is 11.5m wide. You notice a cave on the opposite wall whose ceiling is 6.4m below your feet. The cave is 4.7m deep, and has a vertical back wall. You decide to kick a rock across the ravine into the cave.
1. What initial horizontal velocity must you give the rock so that the rock barely misses the overhang?
2. How far down the back wall of the cave will the projectile hit?



2. delta y = v0yt - 1/2 gt^2
delta x = vox t


3. First I used y = -4.9t^2 to solve for t. After that, I tried to plug the t into the delta x equation to find the initial velocity. But for delta x, I don't know which distance I should use...11.5 or (11.5+4.7)?

Thanks!

 

[PhanthomJay]

1. You are standing on the edge of a ravine that is 11.5m wide. You notice a cave on the opposite wall whose ceiling is 6.4m below your feet. The cave is 4.7m deep, and has a vertical back wall. You decide to kick a rock across the ravine into the cave.
1. What initial horizontal velocity must you give the rock so that the rock barely misses the overhang?
2. How far down the back wall of the cave will the projectile hit?



2. delta y = v0yt - 1/2 gt^2
delta x = vox t


3. First I used y = -4.9t^2 to solve for t. After that, I tried to plug the t into the delta x equation to find the initial velocity. But for delta x, I don't know which distance I should use...11.5 or (11.5+4.7)?

Thanks!

When you solved for t, the rock had descended 6.4m, just missing the overhang. What's its horizontal distance at that instant?

 

[m2003]

Hello,

I would like to provide some clarification and additional information regarding projectile motion and your specific scenario.

1. In order to calculate the initial horizontal velocity, we need to use the formula for horizontal distance, which is given by delta x = v0x * t. Since the rock is being kicked from the edge of the ravine, we can assume that the initial horizontal distance is 11.5m. Therefore, we can use this value for delta x in the equation.

Next, we need to find the time it takes for the rock to reach the other side of the ravine. This can be calculated using the formula for vertical distance, which is given by delta y = v0y * t - 1/2 * g * t^2. In this case, the vertical distance is 6.4m and the acceleration due to gravity, g, is 9.8m/s^2. We can set the final vertical distance to 0, since the rock will be at the same height when it reaches the other side. This gives us the equation 0 = v0y * t - 1/2 * 9.8 * t^2. Solving for t, we get t = 0.9 seconds.

Now, we can plug in this value of t into the horizontal distance equation to find the initial horizontal velocity. This gives us v0x = 11.5 / 0.9 = 12.78 m/s.

2. To calculate how far down the back wall of the cave the projectile will hit, we can use the formula delta y = v0y * t - 1/2 * g * t^2. In this case, we know the initial vertical velocity, v0y, is 0, since the rock is being kicked horizontally. Therefore, the equation becomes delta y = -1/2 * 9.8 * t^2. Plugging in the value of t we calculated earlier, we get delta y = 2.9m. This means that the projectile will hit the back wall of the cave 2.9m below the entrance.

I hope this helps to answer your questions. Please let me know if you have any further inquiries. Thank you.