If the height of ball is shown by h(t)=52t-16[tex]t^{2}[/tex] then the ball will reach after 1.625 seconds.
Given that height in t seconds is shown by h(t)=52t-16[tex]t^{2}[/tex].
We are required to find the time after which the ball will attain its maximum height.
The maximum height is the y coordinate of vertex of the parabola. Then we can use the following value of t.
h(t)=52t-16[tex]t^{2}[/tex]
Differentiate with respect to t.
dh/dt=52-32t
Again differentiate with respect to t.
[tex]d^{2}h/dt^{2}[/tex]=-32t
Because tim cannot be negative means the height is maximum.
Put dh/dt=0
52-32t=0
-32t=-52
t=52/32
t=1.625
Hence if the height of ball is shown by h(t)=52t-16[tex]t^{2}[/tex] then the ball will reach after 1.625 seconds.
Question is incomplete as the right and complete equation is
h(t)=52t-16[tex]t^{2}[/tex] .
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