Note: Use principles of physics to solve the problem and then verify your answer

Question

Note: Use principles of physics to solve the problem and then verify your answer using the simulation.
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 = +4.50 m/s. You should use principles of physics to answer the following questions, and then verify your answers with the simulation.
(a) What is the maximum y-position reached by the ball?
m

(b) At what time does the ball reach its maximum y-position?
s

(c) At what time does the ball pass y = 0?
s

(d) What is the y-velocity when the ball passes y = 0?
m/s

Explanation / Answer

intial velocity= 4.50 m/s
at maximum height velocity= 0, so time taken = (4.50-0)/9.8 s
= 0.459 s

maximum height reached= 0.5+ av*velocity * time
= 0.5+ ( 4.5+0)/2 * 0.459
H2 = 1.532 m

time taken from maximum height to ground = sqrt(2*H2/g) + 0.459

                                                                   =1.01 s

velocity at y=0 = sqrt(2*g*H2)= sqrt(2*9.8*1.532)
=5.47 m/s

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