Please help I will rate! :) Fully explain each step. At time t = 0, a ball is st
ID: 1907456 • Letter: P
Question
Please help I will rate! :) Fully explain each step.
At time t = 0, a ball is struck at ground level and sent over level ground. The figure below gives the magnitude p of the ball's momentum versus time t during the flight. (P1 = 5.0 kg middot m/s and the vertical axis is marked in increments of 1.0 kg-m/s.) At what initial angle above the horizontal is the ball launched? What is the definition of momentum? What is the vertical component of the momentum when the ball reaches maximum height? What, then, is the horizontal component? What is the magnitude of the momentum when the ball is launched? What, then, is the launch angle?Explanation / Answer
We generally do not answer the questions of those who are having low rating. You are having a rating of "96%". Please rate all answers and improve your rating. Even if you find the answers are incorrect, rate them as not helpful. But do rate them for the good of both of us. Thanks!! Coming to the problem Initial momentum = p1+4 = 9 kg m/s =>mv = 9 As the ball reaches maximum height, horizontal velocity remains constant, vertical velocity decreases as there is deceleration in vertical direction due to gravity. So v decreases. So momentum decreases. At maximum height there will be only horizontal speed and vertical speed is zero. So momentum is minimum. So mu = 5 kg m/s => u = 5/m where u is horizontal speed. Let v' is vertical speed =>v = sqrt(v'^2 + u^2) =>v^2 = v'^2 + u^2 =>(9/m)^2 = v'^2 + (5/m)^2 =>v' = (1/m)*sqrt(81-25) = vertical speed Let it is launched at an angle teta =>tan(teta) = v'/u = (1/m)*sqrt(81-25)/5/m = sqrt(81-25)/5 = 1.50 =>teta = 56.25 Degrees