Respuesta :
Y=y1 + v1 + 1/2gt^2
Y= 6m+35m/s^2*t+1/2(9.8m/s^3)t^2
Y= 6m+35m/s^2*t+1/2(9.8m/s^3)t^2
we know that
acceleration due to gravity is 9.8m/s^2
so, we get
[tex] a(t)=9.8 [/tex]
and acceleration will always be constant
we know that
integral of acceleration is velocity
so, we can integrate both sides
[tex] \int \:a\left(t\right)dt=\int \:9.8dt [/tex]
[tex] v(t)=9.8t+C [/tex]
we are given
v(0)=35m/s
we can use it and find C
[tex] v(0)=9.8*0+C [/tex]
[tex] 35=9.8*0+C [/tex]
[tex] C=35 [/tex]
now, we can plug it back
[tex] v(t)=9.8t+35 [/tex]
we know that
integral of velocity is height
we can integrate it again
[tex] \int \:v\left(t\right)dt=\int \:\left(9.8t+35\right)dt [/tex]
[tex] s(t)=\int \:\left(9.8t+35\right)dt [/tex]
[tex] s(t)=4.9t^2+35t+C [/tex]
now, we have
s(0)=6
we can use it and find C
[tex] s(0)=4.9(0)^2+35(0)+C [/tex]
[tex] 6=4.9(0)^2+35(0)+C [/tex]
now, we can plug back C
[tex] C=6 [/tex]
and we get
[tex] s(t)=4.9t^2+35t+6 [/tex].............Answer