Question

A projectile is launched upward with a velocity of 64 feet per second from the top of a 45-foot platform. What is the maximum height attained by the projectile?
height function

Answer 1

Answer:

Step-by-step explanation:

xf=xo+vot+1/2 g t^2

xo=45 f    vo=64 f/s   g = -32.2 ft/s^2

h(t) = -16.1 t^2 + 64t +45        parobola max will be at t = -b/2a

              -64/(2(-16.1) =1.987 sec = tmax

-16.1(1.987^2)+64(1.987)+45 = 108.6 ft