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 v0, 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 Δt on the way up and down.
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- 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.
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- 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.