Question

a cliff diver plunges from a height of 25 ft above the water surface. the distance the diver falls in t seconds is given by the function (d(t)=16t^{2}) ft. which equation can be solved for t to find the time (in seconds) when the diver hits the water?
(16t^{2}-25 = 25)
(16t^{2}+25 = 0)
(16t^{2}+25=-25)
(16t^{2}=25)
(16t^{2}=0)
after how many seconds will the diver hit the water?
given that the velocity of the diver at time (t = a) is given by (lim_{h
ightarrow0}\frac{d(a + h)-d(a)}{h}), what value of a (in s) should be used to calculate the velocity of the diver when they hit the water?
determine the value of (d(a)) (in ft) when the diver hits the water.
with what velocity (in ft/s) does the diver hit the water?

Explanation:

Step1: Find the equation for when the diver hits water

The diver starts 25 ft above water and the distance fallen is $d(t)=16t^{2}$. When the diver hits the water, the distance fallen equals the initial height. So $16t^{2}=25$.

Step2: Solve for $t$

We have the equation $16t^{2}=25$. Divide both sides by 16: $t^{2}=\frac{25}{16}$. Then take the square - root of both sides. Since $t\geq0$ (time cannot be negative in this context), $t = \sqrt{\frac{25}{16}}=\frac{5}{4}=1.25$ s.

Step3: Determine the value of $a$ for velocity calculation

The value of $a$ should be the time when the diver hits the water. So $a = 1.25$ s.

Step4: Find $d(a)$

Since $d(t)=16t^{2}$ and $a = 1.25$, then $d(a)=16\times(1.25)^{2}=16\times\frac{25}{16}=25$ ft.

Step5: Calculate the velocity

First, $d(a + h)=16(a + h)^{2}=16(a^{2}+2ah+h^{2})$ and $d(a)=16a^{2}$. Then $\frac{d(a + h)-d(a)}{h}=\frac{16(a^{2}+2ah + h^{2})-16a^{2}}{h}=\frac{16a^{2}+32ah+16h^{2}-16a^{2}}{h}=\frac{32ah + 16h^{2}}{h}=32a+16h$. Taking the limit as $h
ightarrow0$, we get the velocity $v = 32a$. Substituting $a = 1.25$ s, we have $v=32\times1.25 = 40$ ft/s.

Answer:

• The equation is: $16t^{2}=25$
• $t = 1.25$ s
• $a = 1.25$ s
• $d(a)=25$ ft
• $v = 40$ ft/s