If an object is propelled upward from a height of h feet at an initial velocity of v feet per​ second, then its height S after t seconds is given by the equatio

Question

If an object is propelled upward from a height of h feet at an initial velocity of v feet per second, then its height S after t seconds is given by the equation Sequals=minus−16t squaredt2plus+vtplus+h, where S is in feet. If the object is propelled from a height of 44 feet with an initial velocity of 9696 feet per second, its height S is given by the equation Sequals=minus−16t squaredt2plus+9696tplus+44.
After how many seconds is the height 144144 feet?

Answer 1

Student, 

 

You really need to edit the problem statement.  It looks like all of the numbers are appearing as doubles.  I assume you meant to say

 

s(t) = -16t² + 96t + 4

 

and that you want the time to reach a height of 144 feet.  Set s(t) = 144 and solve for t.

 

144 = -16t² + 96t + 4

 

-16t² + 96t - 140 = 0

 

Solve this quadratic equation for t (which you should know how to do).  You should get 2 solutions:

 

t₁ = 2.5 s

 

This is the time to reach 144 ft on the way up.

 

t₂ = 3.5 s

 

This is the time to reach 144 ft again, on the way down.