Question

A projectile is fired straight upward from ground level with an initial velocity of `128` feet per second. At what instant will it be back at ground level? When will the height be less than `128` feet?

Answer 1

The total flight time will be `7.96` `s`.

The projectile will be lower than `128` `ft` when `t<0.26` `s` and `t>7.7` `s` (until it hits the ground).

Explanation:

I'm used to working in metres, but if we remember that the acceleration due to gravity is `32.2` `fts^-2` we'll be OK.

The simplest way to find the total time is to divide the motion into two sections, the trip up to the highest point and the trip back.

The initial velocity, `u`, is `128` `fts^-1` upward. We'll choose upward as the positive direction. With this choice, the acceleration due to gravity will be negative, since it is in the downward direction.

At the top of the projectile's motion, its velocity, `v`, will be `0` `fts^-1`.

`v=u+at`

Rearranging to find `t`:

`t=(v-u)/a=(0-128)/-32.2=3.98` `s`

Remember, this was half the journey, so the total time for the projectile to return to ground level will be `7.96` `s`.

To find the times (there will be two) when the projectile is less than `128` `ft` above the ground we can use:

`s=ut+1/2at^2`

`128 = 128t+1/2(-32.2)t^2`

`-16.1t^2+128t-128=0`

This is a quadratic equation, which we can solve using the quadratic formula or other methods. It yields:

`t=0.26` `s` or `7.7` `s`

Fascinatingly, the two solutions of the quadratic equation yield the time when the projectile passes though `128` `ft` high on the way up and on the way down!

Answer 2

(1) `t = 7.96` `"s"`

(2) `t in [0, 1.17color(white)(l)"s"]uu[6.79color(white)(l)"s", 7.96color(white)(l)"s"]`

Explanation:

We're asked to find (1) the time when the projectile reaches the ground again, and (2) the time intervals when the height `Deltay` is less than `128` `"ft"`.

(1)

To find the time when the projectile reaches the ground, we can use the equation

`y = y_0 + v_(0y)t + 1/2a_yt^2`

• Since we want the height to be zero, we'll make both the position `y` and initial position `y_0` be zero.

• the initial `y`-velocity is `128` `"ft/s"`

• the acceleration is `-g`, which in these units is

`((9.8cancel("m"))/(1color(white)(l)"s"^2))((3.281color(white)(l)"ft")/(1cancel("m"))) = 32.15` `"ft/s"^2`

Plugging in known values, we have

`0 = 0 + (128"ft/s")t - 1/2(32.15"ft/s"^2)t^2`

`1/2(32.15"ft/s"^2)t^2 = (128"ft/s")t`

`(16.08"ft/s"^2)t = 128"ft/s"`

`t = color(red)(7.96` `color(red)("s"`

(2)

Let's now find the two times when the height `y` is equal to `128` `"m"`. Using the same equation, and plugging in known values, we have

`128color(white)(l)"ft" = 0 + (128"m/s")t - 1/2(32.15"ft/s"^2)t^2`

Forming a quadratic equation,

`(16.08"ft/s"^2)t^2 - (128"ft/s")t + 128color(white)(l)"ft" = 0`

`t = (128+-sqrt((-128)^2-4(16.08)(128)))/(2(16.08)) = color(blue)(1.17` `color(blue)("s"` and `color(blue)(6.79` `color(blue)("s"`

Thus, we can expect that the height of the projectile to be less than `128` `"ft"` during the time intervals

`color(blue)(t in [0, 1.17color(white)(l)"s"]uu[6.79color(white)(l)"s", 7.96color(white)(l)"s"]`

That is, the height of the projectile is less than `128` feet from the launch time until `1.17` seconds, and then again from `6.79` seconds until it hits the ground (at `t = 7.96` `"s"`).

Notice how both intervals last for a duration of `1.17` seconds..