Scenario

A juggler tosses and catches a ball. The ball travels along a vertical line, leaving and returning to the juggler’s hand at the same vertical height. Sketch the ball’s *v-t*, *s-t* and *a-t *graphs.

Solution

- (xmtutor)
- Since this motion involves both upward and downward velocity, the more intuitive “
**upward is positive**” sign convention is adopted.

- The
*v-t* graph starts at *v*_{0}, the velocity at which the ball leaves the juggler’s hand.
- Moving under the influence of gravity only, the velocity decreases linearly at 9.81 m s
^{-2}.
- Notice that the velocity switches sign at the peak since that’s where the velocity transitions from upward to downward direction.
- Notice also the symmetrical nature of the graph, with matching launch and landing speeds, matching traveling distance
*H *and traveling time Δ*t* on the way up and down.

–

- The
*s-t* graph is symmetrical curve reaching the maximum height *H*.
- The graph flattens on the way up as the ball slows down, and steepens on the way down as the ball speeds up.
- By the way, we know the curve is a quadratic one, since this is uniform acceleration motion.

–

- Whether the ball is on its way up or down, or at the peak for that matter, the ball experiences only one force – its own weight. The acceleration is thus constant, 9.81 m s
^{-2}, downward.

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