h = -4.9(t-1)²+6.3

Question

 

A trampoline artist bounces on a trampoline. Her height above the ground is modelled by the function above.

h measures height....t measures time after she leaves the trampoline in seconds.

 

Question 1:how high is the trampoline off the ground? 

Question 2:how long after she leaves the trampoline does she reach her max. height?

 

Answer 1

h(t) = -4.9(t-1)²+6.3

A trampoline artist bounces on a trampoline. Her height above the ground is modelled by the function above.
h measures height [IN METERS]....t measures time after she leaves the trampoline in seconds.

Question 1:how high is the trampoline off the ground?

 

At t = 0 the artist is ON the trampoline.

 

h(0) = -4.9(0-1)²+6.3 = -4.9+6.3 = 1.4 meters off ground.

Question 2:how long after she leaves the trampoline does she reach her max. height?

 

h(t) = -4.9(t-1)²+6.3 is the equation of a parabola with Axis of Symmetry at t = 1. The Vertex is at (1,6.3) and is the maximum of the parabola because it opens down (leading coefficient is negative).

 

She leaves trampoline at t = 0 and reaches maximum height at t = 1, so it took her 1 second to reach maximum height.

Answer 2

1.  At t = 0 find height from the equation
 
h = -4.9(t-1)2 + 6.3
 
h = -4.9(0-1)2 +6.3 = 1.4 units
 
2. After artist  leaves the trampoline its instantaneous velocity keeps reducing and at maximum height instantaneous velocity becomes zero.
 
Instantaneous velocity = differentiation of h with respect to t= dh/dt = -4.9 x 2(t-1) + 0
 
0 = -9.8(t-1)
 0 = -9.8t + 9.8
t= 1 sec
 
Now you can find h at t=1s
 
h = -4.9(1-1)2 + 6.3
h = 6.3 units
:)