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apply 13. the height h in feet that a pole vaulter can reach can be est…

Question

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  1. the height h in feet that a pole vaulter can reach can be estimated using the formula h = \\(\frac{v^{2}}{20}\\), where v is the velocity of the athlete in feet per second. if the height for a womens pole vault is about 9.8 feet, about how fast was the holder running? how far will the pole vaulter run in 2 seconds?
  2. the formula h = 16\\(t^{2}\\) describes the time t in seconds that it takes for an object to fall from a height of h feet. an object is dropped from a height of 1,600 feet. how many seconds does it take to reach the ground?
  3. assess reasonableness deisha says that the equation \\(x^{2}=-25\\) is unable to be solved. do you agree? why or why not?
  4. build perseverance solve the equation 5\\(x^{2}-10 = 70\\).
  5. determine if the following statement is true or false. if false, give a counterexample. the cube root of zero is zero.
  6. build perseverance solve the equation \\(x^{2}=1.21\\).

Explanation:

Response
13.

Step1: Find the velocity

Given $h = \frac{v^{2}}{20}$ and $h = 9.8$. Substitute $h$ into the formula:
$9.8=\frac{v^{2}}{20}$
Multiply both sides by 20: $v^{2}=9.8\times20 = 196$
Take the square - root of both sides: $v=\sqrt{196}=14$ feet per second.

Step2: Find the distance

Using the formula $d = vt$, with $v = 14$ feet per second and $t = 2$ seconds.
$d=14\times2 = 28$ feet.

Given $h = 16t^{2}$ and $h = 1600$. Substitute $h$ into the formula:
$1600=16t^{2}$
Divide both sides by 16: $t^{2}=\frac{1600}{16}=100$
Take the square - root of both sides. Since $t>0$ (time cannot be negative in this context), $t = \sqrt{100}=10$ seconds.

In the set of real numbers, for any real number $x$, $x^{2}\geq0$. The equation $x^{2}=-25$ has no real - number solutions because the square of a real number is always non - negative.

Answer:

The pole - vaulter was running at 14 feet per second and ran 28 feet in 2 seconds.

14.