Question

you throw a water balloon off of your 4th floor apartment balcony with a velocity of 32 feet per second and the balcony is 48 feet high. use the formula $h(t)=-16t^{2}+v_{0}t+h_{0}$, where $h(t)$ is the initial height of the object in feet, $v_{0}$ is the objects initial velocity if feet per second, $h_{0}$ is the objects initial height and $t$ is the time in seconds.
a. how long after the balloon is thrown will it reach the ground?
b. when will the balloon reach its maximum height in the air? what will that height be?

Explanation:

Step1: Identify the formula and given values

The height formula is \(h(t)=-16t^{2}+v_{0}t + h_{0}\). Here, \(v_{0} = 32\) (initial velocity) and \(h_{0}=48\) (initial height).

Step2: Solve part a (when the balloon reaches the ground, \(h(t) = 0\))

Set \(h(t)=0\), so \(-16t^{2}+32t + 48=0\). Divide through by \(-16\) to get \(t^{2}-2t - 3=0\). Factor the quadratic: \((t - 3)(t+ 1)=0\). Using the zero - product property \(t-3 = 0\) or \(t + 1=0\). Since time \(t\geq0\), we discard \(t=-1\).

Step3: Solve part b (find the time of maximum height)

For a quadratic function \(y = ax^{2}+bx + c\) (in our case \(a=-16\), \(b = 32\)), the time \(t\) of the vertex (maximum for \(a<0\)) is given by \(t=-\frac{b}{2a}\). Substitute \(a=-16\) and \(b = 32\) into \(t=-\frac{b}{2a}\), we get \(t=-\frac{32}{2\times(-16)}=1\) second. Then find the maximum height by substituting \(t = 1\) into \(h(t)\): \(h(1)=-16\times(1)^{2}+32\times(1)+48=-16 + 32+48=64\) feet.

Answer:

a. The balloon will reach the ground at \(t = 3\) seconds.
b. The balloon will reach its maximum height at \(t = 1\) second and the maximum height is \(64\) feet.