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A ball is thrown into the air. Its height (in feet) t seconds later is given by:

h(t)=128t-16t^2

Based on the two equivalent forms of the function:

h(t)=128t-16t^2=t*(128-16t), answer the following questions:

a) What is the height of the ball 1 second after it has been thrown?

b) After how many seconds does the ball hit the ground?

c) At what time(s) is the ball 10 feet above the ground? If there is more than one answer, give your answer as a comma separated list of values.

Respuesta :

Answer:

Step-by-step explanation:

Given that a ball is thrown into the air. Its height (in feet) t seconds later is given by:

[tex]h(t)=128t-16t^2[/tex]

a) the height of the ball 1 second after it has been thrown

[tex]= 128(1)-16(1) = 112[/tex]

b) The ball hits the ground when h =0

[tex]h(t)=128t-16t^2[/tex]=0

t=0 or 8

After 8 seconds it hits the ground

c) [tex]10 = 128t-16t^2\\8t^2-64t+10 =0\\t=\frac{64+/-\sqrt{64^2-320} }{16} \\=0.159 seconds, 7.841[/tex]

so after 0.159 seconds and 7.841 seconds it is at height 10 feet

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