h(t)=-16t^2+100t+25

Question

Initial height

How many seconds to reach max

Rocks max height

Seconds to hit the ground 

 

Answer 1

The initial height is the height at time 0, so we plug in 0 for t and solve:

 

h(t) = -16t² + 100t + 25

h(0) = -16(0) + 100(0) + 25

h(0) = 25

 

The initial height at time 0 is 25 (units?).

 

The time to reach maximum height can be represented by finding the vertex of this parabola.  The formula for the x component of the vertex is x = -b/2a.  This will give us the time it takes for the rock to reach the maximum height.

 

x = -b/2a

x = -100/2(-16)

x = 3.125

 

It takes 3.125 (units?) for the rock to reach the maximum height.

 

The maximum height can be found by plugging in the time to reach the maximum height into the equation and solving.

 

h(t) = -16t² + 100t + 25

h(3.125) = -16(3.125)² + 100(3.125) + 25

h(3.125) = -16(9.765625) + 312.5 + 25

h(3.125) = -156.25 + 312.5 + 25

h(3.125) = 181.25 

 

The maximum height is 181.25 (units?).

 

The time it takes to hit the ground can be found by finding the zeroes of the equation or by multiplying the time it takes to reach the maximum height by two because the time it takes to go up should take the same amount of time to come down.  

 

h(t) = -16t² + 100t + 25

0 = -16t² + 100t + 25

-25 = -16t² + 100t

-25 = -4t(4t - 25)

 

-4t = -25

t = 25/4

 

-25 = 4t - 25

0 = 4t

0 = t

 

Our zeroes are t = 0 and t = 25/4, so the total time is 25/4, or 6.25 (units?).