Question

An object is thrown vertically from a height of `5 "m"` at `3 "m/s"`. How long will it take for the object to hit the ground?

Answer 1

`0.7 quad "s"`

Explanation:

"Thrown vertically" could mean either upwards or downwards. I am assuming it to be downwards.

Step 1: Understand the problem.

An object is under free fall. It is accelerating at `g = 9.81 quad "m/s"^{2}` downwards. It is given an initial velocity of `3 quad "m/s"` downwards.

Step 2: Identify the relevant equations.

To find displacement, `s`, under constant acceleration, `a`, and initial velocity, `v_0`, after time, `t`, the following equation is applicable:

`s = v_0 t + 1/2 a t^2`

To get the time as a function of the distance traveled, make `t` as the subject of formula. Begin by writing the quadratic equation in standard form.

`t^2 + frac{2 v_0}{a}t - frac{2 s}{a} = 0`

Subsequently, complete the square.

`(t + frac{v_0}{a})^2 - frac{2 s}{a} = (frac{v_0}{a})^2`

`(t + frac{v_0}{a})^2 = frac{v_0^2}{a^2} + frac{2 s}{a}`

`= frac{v_0^2 + 2as}{a^2}`

Reject the negative square root and take the positive one. This is because we only restrict ourselves to `t > 0`.

`t + frac{v_0}{a} = frac{sqrt{v_0^2 + 2as}}{a}`

`t = frac{sqrt{v_0^2 + 2as} - v_0}{a}`

Step 3: Set up the coordinate system.

I let the downwards direction be positive. The ground level is arbitrarily set to be the origin (i.e. `s = 0`). This is up to your preferences.

Step 4: Identiy the parameters.

In this question,

• `a = g = 9.81 quad "m/s"^2`

• `v_0 = 3 quad "m/s"`

• `s = 5 quad "m"`

Step 5: Plug into the equation.

`t = frac{sqrt{v_0^2 + 2as} - v_0}{a}`

`= frac{sqrt{(3 quad "m/s")^2 + 2(9.81 quad "m/s"^2)(5 quad "m")} - (3 quad "m/s")}{9.81 quad "m/s"^2}`

`= 0.749 quad "s"`

Now earlier, I assume that the object has an initial velocity of `3 quad "m/s"` downwards. The formula also works if the initial velocity is `3 quad "m/s"` upwards. In that case, everything remains the same, except

• `v_0 = - 3 quad "m/s"`

Plugging in the values again

`t = frac{sqrt{v_0^2 + 2as} - v_0}{a}`

`= frac{sqrt{(-3 quad "m/s")^2 + 2(9.81 quad "m/s"^2)(5 quad "m")} - (-3 quad "m/s")}{9.81 quad "m/s"^2}`

`= 1.361 quad "s"`