Question

If an object is launched straight up into the air from a starting height of h_{0} feet, then the height of the object after t seconds is approximately h=-16t^2+v_{0}t+h_{0} feet, where v_{0} is the initial velocity of the object. Find the starting height and initial velocity of an object that attains a maximum height of 412 feet five seconds after being launched.

Answer 1

h=-16t^2+v_{0}t+h_{0}
=> final velocity = h'(t)
h'(t) = -32t + v(0)
Maximum height is achieved after 5 seconds:
=> h'(5) = -32*5 + v(0)
=> v(0) = h'(5) + 32*5 = h'(5) + 160
Remember maximum height => h'(t) = 0. i.e. final velocity must be zero at maximum height just before the object starts decending.
=> v(0) = 160 + 0 = 160

Initial velocity = 160


b) Starting height:

h=-16t^2+v_{0}t+h_{0}
we know that v(0) = 160 ( from above)

=> h=-16t^2+v_{0}t+h_{0}
=> h=-16*25+ 160*5 + h_{0}
=> h =-400+ 800 + h_{0}
But h = 412 ( given)
=> 412 = 400 + h_{0}
=>h_{0} = 12


Initial height =12
Initial velocity = 160

Answer 2

FYI you can prove the above plugging h(0) and v(0) values into the original equation : h=-16t^2+v_{0}t+h_{0}
This should give the value of h as 412 concurring with the value given in the question.

The trick was:

a) differentiate the original equation to give the final velocity.
b) set that final velocity to zero to get the original velocity.

Answer 3

Thank you very much!

Answer 4

vw