Recent content by tremain74

At first I was going to use the values 7m /s to 99 m/s as the numbers to plug into the equations to solve for the distance but these values are for velocity and not for seconds.

My solution: v = 2*t + 5*t^2. dx/dt = 2*t + 5*t^2. Therefore dx = (2t + 5t^2)dt. After taking the anti derivative my equation is s = t^2/2 + 5/3(t^3).

It's 10 meters per second

I actually forgot that I worked on this problem from last year. But I now remember.

I am back. I used the formula s = 1/2(v+v0)*t. The first was s0 = 1/2*(10+0)*1. The answer was 5. My logic was at first the stone left the balloon starting the velocity as 0. Then I continued to use the formula s1 =1/2 (v+v0)*2 for two seconds and ect. every time I used the formula I add a...

I am back. I used the formula s = 1/2(v+v0)*t. The first was s0 = 1/2*(10+0)*1. The answer was 5. My logic was at first the stone left the balloon starting the velocity as 0. Then I continued to use the formula s1 =1/2 (v+v0)*2 for two seconds and ect. every time I used the formula I add a...

TL;DR Summary: I have a problem that I am working on in my Engineering Dynamics Book. A stone is dropped from a balloon that is ascending at a uniform rate of 10m/s. If it takes the stone 10 seconds to reach the ground, how high was the balloon at the instant the stone was dropped? The answer...

This is my solution below.

The one formula that I thought about was the projectile motion formula using y = -4.9t^2 + vo*t*sinO. Sin O(sin of theta). However this does not solve my problem to find the correct height.

To find the time of this problem, I used the equations 3 + (10 * x ) = 13.5 *x. 3 = 3.5*x. Therefore x was 0.857. To start off with child A was already three feet above ground and I got the answer to the time. However I couldn't find the height.

My free body diagram doesn't include the ice that comes as a result of water that will eventually freeze. However, in reality, I definitely have to include the ice in the fbd as well.

Actually, I meant that the dry ice is converted into carbon dioxide. My apology.

This is my attached picture.

I have an object that is going up the plane at 25 degrees to the right. It is going at 15 m/s. I drew a free body diagram of the object. I have a P force at 15 m/ s going up the direction at 25 degrees. I also have friction force going in the opposite direction of 0.25 * N.

I have tried to used the drawing pad last night to show my work but I am having trouble using it.