time to max ... 224 / 32 ... seconds
average velocity ... 224 / 2
velocity * time = height above platform
remember to add in the platform height
An arrow is launched upward with a velocity of 224
feet per second from the top of a 90
-foot platform. What is the maximum height attained by the arrow?
2 answers
height = -16t^2 + 224t + 90
completing the square:
height = -16(t^2 - 14t + ....) + 90
= -16t^2(t^2 - 14t + 49 - 49) + 90
= -16(t-7)^2 + 784+90
= -16(t-7)^2 + 874
so the arrow will reach a max height of 874 ft after 7 seconds
or, using Calculus
d(height)/dt = -32t + 224 = 0 for a max of height
32t = 224
t = 7
sub 7 into height = -16t^2 + 224t + 90
= -16(49) + 224(7( + 90
= 874
completing the square:
height = -16(t^2 - 14t + ....) + 90
= -16t^2(t^2 - 14t + 49 - 49) + 90
= -16(t-7)^2 + 784+90
= -16(t-7)^2 + 874
so the arrow will reach a max height of 874 ft after 7 seconds
or, using Calculus
d(height)/dt = -32t + 224 = 0 for a max of height
32t = 224
t = 7
sub 7 into height = -16t^2 + 224t + 90
= -16(49) + 224(7( + 90
= 874