A community for students.
Here's the question you clicked on:
 0 viewing
anonymous
 4 years ago
show that time of ascent=time of descent
anonymous
 4 years ago
show that time of ascent=time of descent

This Question is Open

anonymous
 4 years ago
Best ResponseYou've already chosen the best response.0when the velocity of a body increses air resistance increases and so the weigt decreases

anonymous
 4 years ago
Best ResponseYou've already chosen the best response.0set the derivative of vertical location with respect to time to zero and solve for time. This gives you the time when speed is zero, which is when the object stops ascending and begins descending. Compare that result with vertical location = 0, which has two solutions: 0, when the object is released and the time when the object returns to the ground. Example: YY0 = V0*sinA*time_bottom  (g/2)*time_bottom^2 = 0 Add the acceleration term to get: V0*sinA*time_bottom = (g/2)*time_bottom^2 Divide by t (t is not zero, because you don't care for that solution, so division is allowed) V0*sinA = (g/2)*time_bottom divide by (g/2) to get an expression for t time_bottom = 2*V0*sinA/g now take the derivative of the initial with respect to time and set to zero (velocity in the vertical direction is zero when the object is at it's peak) V0*sinA  2*(g/2)*time_top = 0 a little manipulation yields: V0*sinA/g = time_top Combine the two equations to show that: time_top * 2 = time_bottom

anonymous
 4 years ago
Best ResponseYou've already chosen the best response.0Let the initial velocity be u During ascent: acceleration = g Ascent completes when its velocity becomes zero. (=> final velocity v = 0) Using v=u+at, 0 = u + (g)t = u gt Therefore, t (time of ascent) = u/g And height to which the object ascends is given by s = ut + (1/2)at^2 = u(u/g) + (1/2)(g)(u*u)/g*g = u*u/g  u*u/2g = u*u/2g During descent, initial velocity = 0, the distance it has to descend = u*u/2g, and the acceleration = g again using s = ut + (1/2)(at*t), u*u/2g = 0 + (1/2)(g*t*t) => g*t*t = u*u/g => t (time of descent) = u/g Therefore time of ascent = time of descent
Ask your own question
Sign UpFind more explanations on OpenStudy
Your question is ready. Sign up for free to start getting answers.
spraguer
(Moderator)
5
→ View Detailed Profile
is replying to Can someone tell me what button the professor is hitting...
23
 Teamwork 19 Teammate
 Problem Solving 19 Hero
 Engagement 19 Mad Hatter
 You have blocked this person.
 ✔ You're a fan Checking fan status...
Thanks for being so helpful in mathematics. If you are getting quality help, make sure you spread the word about OpenStudy.