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.
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...
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.
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.