11/6/2023 0 Comments Flat red bouncing ball silhouetteSo for the gradient we have +v/+t for +9.8m/s. So as time progresses forward, the ball picks up speed in the positive direction, the direction the velocity is in. So velocity starts from 'vi' (the velocity the ball is dropped at), and begins to increase in the positive direction (down). You can start a ball with any velocity at all, under gravitational force the velocity will change to become more positive giving it a positive slope. The gravitational force doesn't care what the velocity is, it will always change the velocity in the positive direction creating a positive acceleration and a positie slope in the v-t graph. The gravitational acceleration is always in the same direction, the positive direction as you have it set up. The slope of the v-t graph is acceleration. Then it jumps back to +9.8 because the object continues to accelerate in the positive (down) direction after the bounce. First, the acceleration will be +9.8 then it jumps up to infinity (for an instantaneous change of momentum for the bounce) or a very high negative value for a slower change of momentum. The acceleration due to gravity is always in the positive direction regardless of the value of the velocity. ![]() ![]() Your acceleration diagram is not correct.
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