Recent content by hannsparks

Figure 2 below shows water at a dam. The depth of the water at the dam is 20 meters. The area of the wall of the dam which is in contact with the water is 500 m 2. The atmospheric pressure is 1 atmosphere and the mass per unit volume density of the water is 1000 kg/m^3...

wow! thanks fermat for all your patience with me I greatly appreciate it! I only discovered this website two days ago and I thank God I did you've been a tremendous help. thanks again

would it make sense to solve for vi using the conservation of energy equation?

no that doesn't make any sense because the speed would be the same as the final speed, so we have to calculate the vi some how

so it would be 10^2+ 0^2=speed^2

a2 + b2 = c2?

ok last part the velocity right before he hits the water vi=10 vf=? a=9.8 delta x=7 delta y= 2.4 im i on the right track?

ok 490yi=25(10^2)+1176 yi=7.5m

ok so 7/0.6998 which equals 10m/s is the velocity as he leaves the slide

ok so the 0.6998 can be used to find the initial height

im stuck don't know what to do next

no i have to use the time it took him to hit the water to find the initial height I would use another kinematic equation but which one

ok then the time for the horizontal would be 1.19522

deltay = voy t + 1/2 ay t^2 y direction deltax = vox t + 1/2 ax t^2 x direction did i chose the right equations

deltay = voy t + 1/2 ay t^2 is it this one? where t=0.6998