Question

The height of a stuntperson jumping off a building that is 20 m high is modeled by the equation h = 20 -57, where t is the time

Answer 1

A stuntman jumping off a 20-m-high building is modeled by the equation h = 20 – 5t2, where t is the time in seconds. A high-speed camera is ready to film him between 15 m and 10 m above the ground. For which interval of time should the camera film him?

Answer:

$1\leq t\geq \sqrt{2}$

Step-by-step explanation:

Given:

A stuntman jumping off a 20-m-high building is modeled by the equation

$h =20-5t^{2}$-----------(1)

A high-speed camera is ready to making film between 15 m and 10 m above the ground

when the stuntman is 15m above the ground.

height $h = 15m$  

Put height value in equation 1

$15 =20-5t^{2}$

$5t^{2} =20-15$

$5t^{2} =5$

$t^{2} =1$

$t =\pm1$

We know that the time is always positive, therefore $t=1$

when the stuntman is 10m above the ground.

height $h = 10m$  

Put height value in equation 1

$10 =20-5t^{2}$

$5t^{2} =20-10$

$5t^{2} =10$

$t^{2} =\frac{10}{5}$

$t^{2} =2$

$t=\pm\sqrt{2}$

$t=\sqrt{2}$

Therefore ,time interval of camera film him is $1\leq t\geq \sqrt{2}$