The height in meters of a projectile is modeled the function h(t)= -5t^2+25, where t is the time in seconds. a) Find the point when the object hits the ground. b) Find the average rate of change from the point when the projectile is launched to the point in which it hits the ground. c) Estimate the object’s speed at the point of impact.



Answer :

[tex]h(t)=-5t^2+25\\-5t^2+25=0\\25=5t^2\\5=t^2\\\sqrt{5}=t\\\\A(t)=\frac{f(t+h)-f(t)}{h}\\A(0)=\frac{f(\sqrt{5})-f(0)}{\sqrt{5}}\\A(0)=\frac{0+25}{\sqrt{5}}\\A(0)=\frac{25\sqrt{5}}{5}\\A(0)=5\sqrt{5}\\\\h'(t)=-10t\\h'(\sqrt{5})=-10(\sqrt{5})[/tex]

A. sqrt(5)
B. 5 sqrt(5)
C. 10 sqrt(5)

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