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A ball thrown upwards hits a roof and returns back to the ground.

The upward movement is modeled by a function [tex]s=-t^2+3t+4[/tex]
s= −(t^2)+3t+4
and the downward movement is modeled by [tex]s=-t^2+3t+4[/tex]
s= −2(t^2)+t+7, where s is the distance (in metres) from the ground and t is the time in seconds.

Find the height of the roof from the ground.

Answer :

caylus

Answer: 6 m

A ball thrown upwards from the altitude 4 m,

hits a roof and returns back to the ground.

upward movement:  s= −t²+3t+4

downward movement: s=-2t²+t+7

Step-by-step explanation:

Let's calculate the intersection:

[tex]- t^2+3t+4 =-2t^2+t+7\\\\t^2+2t-3=0\\\\t^2+3t-t-3=0\\\\t(t+3)-(t+3)=0\\\\(t+3)(t-1)=0\\\\t=-3 \ (exclude)\ or\ t=1\\\\if\ t=1 \ then\ s=-1^2+3*1+4=6\\\\height\ is\ 6\ m.\\[/tex]

Sorry, i have forgotten the picture.

${teks-lihat-gambar} caylus

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