Question

A cliff diver plunges from a height of\(100\;ft\)above the water surface. The distance the diver falls in\(t\)seconds is given by the function\(d\left( t \right) = 16{t^2}\;ft\).

(a)After how many seconds will the diver hit the water?

(b) With what velocity does the diver hit the water?

Short answer

(a). After \(2.5\) seconds diver hit the water.

(b). The diver hit the water with velocity \(80\;{\rm{ft/s}}\).

Step-by-step solution

(a) Step 1: Write the time in seconds whendriver hit the car

As the height of the cliff at which the diver plunges is\(100\;{\rm{ft}}\)above the water surface. Distance is\(d\left( t \right) = 16{t^2}\;{\rm{ft}}\).

As the driver hit the water at\(d\left( t \right) = 100\).

Now find time as:

\(\begin{aligned}100 &= 16{t^2}\\t &= \frac{5}{2}\\ &= 2.5\;{\rm{seconds}}\end{aligned}\)

Thus, after \(2.5\) seconds diver hit the water.

(b) Step 2: Find the velocity at which the diver hit the car

Apply the definition ofinstantaneous velocitywhich states that the position function \(f\left( t \right)\) at time \(t = a\) is represented as \(v\left( a \right) = \mathop {\lim }\limits_{h \to 0} \frac{{f\left( {a + h} \right) - f\left( a \right)}}{h}\).

Now find the velocity at time \(t = 2.5\;{\rm{seconds}}\).

\(\begin{aligned}v\left( t \right) &= \mathop {\lim }\limits_{h \to 0} \frac{{d\left( {2.5 + h} \right) - d\left( {2.5} \right)}}{h}\\ &= \mathop {\lim }\limits_{h \to 0} \frac{{16{{\left( {2.5 + h} \right)}^2} - 100}}{h}\\ &= \mathop {\lim }\limits_{h \to 0} \frac{{100 + 80h + {{16}^2} - 100}}{h}\\ &= \mathop {\lim }\limits_{h \to 0} \frac{{80h + 16{h^2}}}{h}\\ &= \mathop {\lim }\limits_{h \to 0} \frac{{h\left( {80 + 16h} \right)}}{h}\\ &= \mathop {\lim }\limits_{h \to 0} \left( {80 + 16h} \right)\\ &= 80\end{aligned}\)

Thus, diver hit the water with velocity \(80\;{\rm{ft/s}}\).