An object is propelled from the ground with an initial velocity of 80 feet per second. How long does it take to reach a height of 100 feet?

Question

I have to write an equation for this problem and solve fir the given information. 

Answer 1

First we have the equation for the height of an object thrown upward:  h(t)=-16t² + v₀t.  Substituting 100 in for h(t), and 80 in for v₀, we have:

100=-16t²+80t

Subtracting 100 from both sides, we end up with

-16t²+80t-100=0

Multiplying both sides by -4 to make the first coefficient a positive number (& smaller) we have

4t²-20t+25=0

Option 1: This can be factored into (2t-5)(2t-5)=0, from which we conclude 2t-5=0, so t=52=2.5 sec.

Option 2: Use vertex t= -b/2a .....so t= -(-20)/2(4) = 20/8=2.5 seconds

 

Answer 2

h = (-1/2)gt² + v₀t + d₀

 

where:

h = final height

g = acceleration due to gravity = 32 ft / s

v₀ = initial velocity = 80 ft/sec

d₀ = initial height = 0 ft

 

h = -16t² + 80t

 

 

Solve for t when h=100.

 

100 = -16t² + 80t

 

 

Subtract 100 to both sides of the equation.

 

0 = -16t² + 80t - 100

 

0 = -2(8t² - 40t + 50)

 

8t² - 40t + 50 = 0

 

 

Use the quadratic formula to solve for t.  Then chose the positive value of t.