Recent content by kman2027

I appreciate your help!

i think i got the initial height (yo) and distance traveled (y) mixed up does this look any better? .3=0+2.4t+1/2(-9.8)t^2 -4.9t^2+2.4t-.3=0 t=.28?

I did: 0=.3+2.4t+1/2(-9.8)t^2 t=.59 Is that the final answer or do you multiply t by 2?

My bad, I had a typo in the equation. .8-1.19t+(1/2)(-9.8)t^2=0 t= .3s Then: (4.44)(.3)= 1.3m Is 1.3m it? Thanks for your help by the way!

Ok, I figured out the first part. The answer is 2.4m/s. The second part has me confused. I believe the question is asking how long the ball is suspended in air from the time the cart launches it till the time the cart catches it, but I'm not sure. Estimate the time interval between...

I got 1.7m this time. I picked the constant acceleration equation: y=yo+voyt-1/2gt^2 plugged in 0=.8+1.19t-1/2(-9.8)t^2 t=.385 (.385)(4.44)=1.7m Is 1.7m right?

A ball is thrown with an initial speed of 4.6m/s at an angle of 15(degrees) below the horizontal. It is released 0.80m above the floor. What horizontal distance does the ball cover before bouncing? Is the answer 2.98? Here's what I did: 4.6sin(15) = initial vertical velocity = 1.19m/s...

That is all the text that is in the question, but there is a picture. If you google image "cart launching a ball," the first picture that comes up is it.

A cart launches a ball 15cm high. The vertical distance from the lowest ball to the highest is about two cart heights or 30cm. Estimate the launch speed of the ball? Estimate the time interval between successive stroboscopic exposures? I honestly have no idea where to begin...

A ball is thrown with an initial speed of 4.6m/s at an angle of 15(degrees) below the horizontal. It is released 0.80m above the floor. What horizontal distance does the ball cover before bouncing? I keep getting 2.4m as my answer, but it's wrong! Here's what I did (Please correct...