An arrow is launched upward with a velocity of 256 feet per second from the top of a 100-foot platform. What is the maximum height attained by the arrow? height function

Question

An arrow is launched upward with a velocity of 256 feet per second from the top of a 100-foot platform. What is the maximum height attained by the arrow?

height function

Answer 1

h(t) = (a/2)t^2 + vot +ho

where vo= initial velocity, ho=initial height

-16t^2=(a/2)t^2 is a term reflecting the effect of gravity's negative acceleration = a=-32 feet per second per second.

h(t) = -16t^2 +256t +100

take the derivative and set = zero

h'(t) = -32t +256 = 0

t = 256/32 = 8 seconds to reach max height

h(8) = -16(8)^2 +256(8) +100

= 16(64) +100 = 1024+100

=1124 feet = maximum height

or if you want to solve it with "just" algebra and no calculus, rewrite the height function in vertex form, graph it, or use an online equation or graphing calculator

h(t) = -16t^2 +256t + 100

= -16(t^2 - 16t + 64) +100+ 16(64)=100+1024

= -16(t-8)^2 +1124

= a(t-h)^2 +k where (h,k)=the vertex

vertex is maximum point = (8,1124) with 1124 ft = max height

graphically, the height function is a downward opening parabola with axis of symmetry x=8 and vertex (8, 1124)=max point