Recent content by AshNotKetchum

I suppose stranger things have happened. Thank you very much ehild.

So, I may have misunderstood you... When I put the expression as as sinθ=(2S*t^2)/g I got the correct answer, which is a horrible way to go about things. Since our kinematics equation is S=0.5gsinθt^2, I don't understand how t^2 came to be in the numerator to get the correct answer.

That is indeed what I meant, sorry for my shoddy mathematical expression. sinθ=(2S)/(g*t^2) In this case, θ is 0.15° while the answer to the question is 0.39°.

Thank you for your reply, ehild and I apologize for not replying much sooner with a result. That gif was quite helpful in visualizing the situation and I thought I understood the problem. Here is what I did: Find time taken to for the travel by Speed/distance. Use the kinematics...

Homework Statement A physics student playing with an air hockey table (a frictionless surface) finds that if she gives the puck a velocity of 3.44 m/s along the length (2.73 m) of the table at one end, by the time it has reached the other end the puck has drifted 2.08 cm to the right but...

Alright, so his speed just before he touches the ground is 9.41 m/s. Then, his acceleration (or deceleration?) is 50.3 m/s^2. Now, given that I have - okay, I think I got it. The answer was supposed to be something in the order of 5388 N and I got 5386 N, rounding it up. Thank you so very...

Well, since the question hinted to assume acceleration is constant while slowing down, I assumed a=g, but at rest a=0, would it not? From what I can understand, there are two forces on the man - gravity (excuse the poor terminology) and the normal force.

Homework Statement A 89.6 kg man steps off a platform 4.52 m above the ground. He keeps his legs straight as he falls, but at the moment his feet touch the ground his knees begin to bend, and, treated as a particle, he moves an additional 0.88 m before coming to rest. Treating our rigid...