SSeshadri
contestada

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?

Respuesta :

chloebunyea

Answer:

−16(2−−5)

Step-by-step 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)

We need to differentiate h(t)

[tex]\\ \sf\longmapsto \dfrac{d}{dt}(-16t^2+16t+80)[/tex]

[tex]\\ \sf\longmapsto \dfrac{d}{dt}(-16t^2)+\dfrac{d}{dt}16t+\dfrac{d}{dt}80[/tex]

[tex]\\ \sf\longmapsto -32t+16[/tex]

Let it be 0

[tex]\\ \sf\longmapsto -32t+16=0[/tex]

[tex]\\ \sf\longmapsto -32t=-16[/tex]

[tex]\\ \sf\longmapsto t=1/2s[/tex]

  • t=0.5s

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