Question

A ball is shot vertically upward from the surface of another planet. A plot of y versus t for the ball is shown in the figure, where y is the height of the ball above its starting point and t = 0 at the instant the ball is shot ?

Answer 1

Free-fall acceleration: `"5.6 m/s"""^2`
Initial velocity: `"14 m/s"`

Explanation:

FULL QUESTION

A ball is shot vertically upward from the surface of another planet. A plot of y versus t for the ball is shown in the figure, where y is the height of the ball above its starting point and t = 0 at the instant the ball is shot.

The figure's vertical scaling is set by `y_s = "21.0 m"`. What are the magnitudes of

(a) the free-fall acceleration on the planet, and
(b) the initial velocity of the ball?

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The question provides you with some very important information, so make sure that you don't overlook anything.

The first important thing to realize is that the ball is being thrown upward, which means that its movement will be one-dimensional, limited to the vertical axis.

This means that the nitial velocity of the ball will be equal to the velocity it has when it return to ground level.

The ball is being decelerated by the gravitational acceleration on its way to the top of its movement, where its velocity is zero, and accelerated by the same gravitational acceleration on its way back to the ground.

So, look at the graph and determine the maximum height reached by the ball. Since you know that `y_s = "21.0 m"`, and that you have six units on the vetical axis, you can say that

`"one unit" = 21.0/6 = "3.5 m"`

Since the ball reaches its maximum height of five vertical units, one shy of `y_s`, you know that

`y_"max" = y_s - 3.5 = "17.5 m"`

At maximum height, the ball's velocity is equal to zero. This means that from the time it got to `y_"max"`, to the time it reaches ground level again, it will be in free fall.

You can tell from the graph that it takes the ball `"2.5 s"` to reach `y_"max"`, and `"2.5 s"` to get back down. From the top of its movement, where `v_"top"=0`, you have

`y_"max" = underbrace(v_"top")_(color(blue)(=0)) * t_"down" + 1/2 * a * t_"down"^2`

`y_"max" = 1/2 * a * t_"down"^2 implies a = (2 * y_"max")/t^2`

Therefore,

`a = (2 * 17.5color(red)(cancel(color(black)("m"))))/(2.5^2"s"^2) = color(green)("5.6 m/s"""^2`

To ge its initial velocity, use the fact that at maximum height its velocity is zero

`underbrace(v_"top")_(color(blue)(=0)) = v_0 - a * t_"up"`

`v_0 = a * t_"up" = 5.6color(red)(cancel(color(black)("m")))/"s"^color(red)(cancel(color(black)(2))) * 2.5color(red)(cancel(color(black)("s"))) = color(green)("14 m/s")`