Question
a projectile is fired upward from the ground to reach 64 feet. What is the velocity when it is 48 feet above the ground
Answers
Steve
y = vt - 16t^2
max height reached when t = v/32
64 = v(v/32) - 16(v/32)^2
64 = v^2/32 - v^2/64
64 = v^2/64
v = 64
y = 64t - 16t^2
max height when t=2
y(2) = 64*2 - 16*4 = 64 as desired.
y' = 64 - 32t
y=48 when t=1 or 3 (going up, coming down)
y'(1) = 32
y'(3) = -32
max height reached when t = v/32
64 = v(v/32) - 16(v/32)^2
64 = v^2/32 - v^2/64
64 = v^2/64
v = 64
y = 64t - 16t^2
max height when t=2
y(2) = 64*2 - 16*4 = 64 as desired.
y' = 64 - 32t
y=48 when t=1 or 3 (going up, coming down)
y'(1) = 32
y'(3) = -32