Question 1118286
The height of the gum can be given by {{{h=65-4.9t^2}}} , where t is in seconds and h is measured in metres.
 
a. To find the average velocity of the gum on the intervals {{{2 <= t <=3}}} and {{{2 <= t <=2.1}}} 
you calculate height of the gum at t=2, t=2.1, and t=3,
and use those values to calculate average velocity.
 
For {{{t=2}}} , {{{h=65-4.9*2^2=65-4.9*4=65-19.6=45.4}}} .
For {{{t=2.1}}} , {{{h=65-4.9*2.1^2=65-4.9*4.41=65-21.609=43.391}}} .
For {{{t=3}}} , {{{h=65-4.9*3^2=65-4.9*9=65-44.1=20.9}}} .
 
In the interval {{{2 <= t <=3}}} , the height of the gum changes from 45.4 m to 20.9 m in 1 second.
The average downwards velocity in the interval {{{2 <= t <=2.1}}} is
{{{(45.4m-20.9m)/(3s-2s)}}}{{{"="}}}{{{24.5m/"1 s"}}}{{{"="}}}{{{"24.5 m / s"}}} .
 
In the interval {{{2 <= t <=2.1}}} , the height of the gum changes from 45.4 m to 43.391 m in 0.1 second.
The average downwards velocity in the interval {{{2 <= t <=2.1}}} is
{{{(45.4m-43.391m)/(2.1s-2s)}}}{{{"="}}}{{{2.009m/"0.1 s"}}}{{{"="}}}{{{"20.09 m / s"}}}
 
b. Knowing about derivatives, you would know that the instantaneous velocity when t=2 is the value of the derivative of h with respect to t when t=2.
The derivative of the function {{{h=65-4.9t^2}}} with respect to t is
{{{dh/dt=-4.9*2t=-9.8t}}} , and its value for {{{t=2}}} is {{{-9.8*2=-19.6}}} .
The derivative is negative because height decreases with time,
but you could say the instantaneous downwards velocity when t=2 is {{{"19.6 m / s"}}} .
If you have not been taught about derivatives as functions, maybe you were expected to calculate it as a limit:
{{{lim(t->2,(h(t)-h(2))/(t-2))}}}{{{"="}}}{{{lim(t->2,(65-4.9t^2-(65-4.9*4))/(t-2))}}}{{{"="}}}{{{lim(t->2,-4.9(t^2-4)/(t-2))}}}{{{"="}}}{{{lim(t->2,-4.9(t+2)(t-2)/(t-2))}}}{{{"="}}}{{{lim(t->2,-4.9(t+2))}}}{{{"="}}}{{{-4.9(2+2))}}}{{{"="}}}{{{-4.9*4}}}{{{"="}}}{{{-19.6}}} .