Question

Hi, I need help with this question and I need an explanation pls. Please don't copy the answer from the internet.

Here's the question: A ball is thrown into the air from the top of a building. The height, h(t), of the ball above the ground t seconds after it is known can be modeled by
h(t) =-16t²+16t+80. How many seconds after being thrown will the ball hit the ground?

Answer 1

We need to differentiate h(t)

`\\ \sf\longmapsto (d)/(dt)(-16t^2+16t+80)`

`\\ \sf\longmapsto (d)/(dt)(-16t^2)+(d)/(dt)16t+(d)/(dt)80`

`\\ \sf\longmapsto -32t+16`

Let it be 0

`\\ \sf\longmapsto -32t+16=0`

`\\ \sf\longmapsto -32t=-16`

`\\ \sf\longmapsto t=1/2s`

• t=0.5s

Answer 2

−16(2−−5)

Explanation:

hay un factor común entre estos números, por lo que simplemente −16t2+16t+80 divida y reutilice y luego reciba−162+16+80 llegar

−16(2−−5)