Respuesta :
Step-by-step explanation:
A ball is launched straight up in the air from a height of 6 feet. The velocity as a function of time t is given by :
f(t) = -32 t+285
Height of the ball is :
[tex]h(t)=\int\limits{f(t){\cdot} dt}\\\\h(t)=\int\limits{(-32t+285){\cdot} dt}\\\\h(t)=-16t^2+285t+C[/tex]
C is constant. Here the ball is launched from a height of 6 feet. So,
[tex]h(t)=-16t^2+285t+6[/tex]
At t = 2 s, [tex]h(t)=-16(2)^2+285(2)+6=512\ m[/tex]
At t = 9 s, [tex]h(t)=-16(9)^2+285(9)+6=1275\ m[/tex]
Between 2 s and 9 s, the ball's height changed is : 1275 - 512 = 763 m.
In between 2 second and 9 second , ball height is changed by 763 feet.
The velocity function is given by,
[tex]f(t)=-32t+285[/tex]
Velocity is defined as rate of change of displacement (or height) with respect to time.
[tex]f(t)=\frac{dh}{dt} \\\\dh=f(t)dt[/tex]
Taking integration on both side.
[tex]h(t)=\int\limits {(-32t+285)} \, dt \\\\h(t)=-32\frac{t^{2} }{2}+285t\\\\h(t)=-16t^{2}+285t[/tex]
Since, ball is launched straight up in the air from a height of 6 feet
So, [tex]h(t)=-16t^{2}+285t+6[/tex]
We have to find ball height changed between 2 s and 9s.
Thus, [tex]h(2)=-16(2)^{2}+285(2)+6=512feet\\\\h(9)= 16(9)^{2}+285(9)+6=1275feet[/tex]
Height change, [tex]= h(9) - h(2)[/tex]
[tex]= 1275 - 512 = 763 feet.[/tex]
Therefore, In between 2 second and 9 second , ball height is changed by 763 feet.
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