A ball is thrown directly upward from an initial height of 100 ft with an initial velocity of 80 ft/sec. After t seconds , the height of the ball above ground

maximum-height
initial-height

#1

A ball is thrown directly upward from an initial height of 100 ft with an initial velocity of 80 ft/sec. After t seconds , the height of the ball above ground is h(t)=-16t^2+80t+100.

after how many seconds does the bal reach its maximum height? what is the maximal height?

after how many seconds will the ball hit the ground?

Answer:

s=ut+1/2at^2 +constant= -16t^2+80t+100. Therefore (1/2)a=-16 Therefore a=32ft/s^2
From parabola theory max occurs at -b/2a = 80/(2*16)=2.5s Therfor 2.5s to reach top. Subst 2.5s in given eq. Think answer is 200ft(check my arith) Now can use s= 200,u=o(at top) with s=ut+1/2at^2 as the eq to use. Therefore 200=(1/2)(32)t^2 Can now find t and add it to the 2.5s