Turn air resistance off so the acceleration is a = -9.80 m/s2 (the - sign indica

Question

Turn air resistance off so the acceleration is a = -9.80 m/s2 (the - sign indicates that the acceleration is down). Set the initial position to y0 = +0.500 m and the initial velocity to v0y = +3.00 m/s. You should use principles of physics to answer the following questions, and then verify your answers with the simulation.
What is the maximum y-position reached by the ball?
At what time does the ball reach its maximum y-position? I found this it was 0.31s.
At what time does the ball pass y = 0?
What is the y-velocity when the ball passes y = 0?

Explanation / Answer

intial velocity= 3 at maximum height velocity= 0, so time taken = (3-0)/9.8 s = .306 s maximum height reached= 0.5+ av*velocity * time = 0.5+ ( 3+0)/2 * 0.306 H2=.959m time taken from maximum height to ground = sqrt(2*H2/g) + 0.306=.44+.306=.748s velocity at y=0 = sqrt(2*g*H2)= sqrt(2*9.8*.959) =4.33 m/s

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