Question

A projectile is launched from ground level with an initial velocity of vo feet per second. Neglecting air resistance, its height in feet t seconds after launch is given bys= - 16t^2+Vot. Find the time(s) that the projectile will (a) reach a height of 272 ft and (b) return to the ground when vo is 32 feet per second. Why is the answer and why?

Answer 1

We are given that the height of an object that is moving in projectile motion is the following:

`S=-16t^2+v_0t`

Where:

`\begin{gathered} S=\text{ height} \\ t=\text{ time} \\ v_0=\text{ initial velocity} \end{gathered}`

Now we will substitute the value of S for 272 ft, we get:

`272=-16t^2+v_0t`

Now we subtract 272 from both sides, we get:

`-16t^2+v_0t-272=0`

We get an equation of the form:

`at^2+bt+c=0`

The value of "t" are determined using the quadratic formula:

`t=\frac{-b\pm\sqrt[]{b^2-4ac}}{2a}`

Substituting the values we get:

`t=\frac{-v_0\pm\sqrt[]{v^2_0-4(-16)(-272)}}{2(-16)}`

Solving the operations we get:

`t=\frac{-v_0\pm\sqrt[]{v^2_0-17408}}{-32}`

Therefore, we get two possible values for the time:

`t_1=\frac{-v_0+\sqrt[]{v^2_0-17408}}{-32}`

The second value is:

`t_2=\frac{-v_0-\sqrt[]{v^2_0-17408}}{-32}`

Part b. We are asked to determine the time when the object returns to the ground and the initial velocity is 32 ft/s. To do that we will use the formula for the height:

`S=-16t^2+v_0t`

Now we substitute the value of the initial speed:

`S=-16t^2+32t`

Now, since we want the time when the object returns to the ground this means that the height must be zero, therefore, we substitute the values S = 0, we get:

`0=-16t^2+32t`

Now we take "t" as a common factor:

`0=t(-16t+32)`

Now we set each factor to zero:

`t=0`

For the second factor we get:

`-16t+32=0`

Now we subtract 32 from both sides:

`-16t=-32`

Now we divide both sides by -16:

`t=-(32)/(-16)`

Solving the operations:

`t=2`

Therefore, in 2 seconds the object will return to the ground.