calculus 1 question

Question

If a ball is thrown straight up into the air with an initial velocity of 65 ft/s, it height in feet after t second is given by y = 50t −16t 2 . Find the average velocity for the time period beginning when t=2 and lasting

Answer 1

for initial velocity of vo = 65 ft/s

the equation for height, with zero intitial height is

y= h(t) = -16t^2 +65t, not -16t^2 +50t

velocity at t=2 is

v(h'(t) = -32(2)+65

= -64+65

= 1 ft/sec

average velocity

= average of velocity at time t=2 to to time t

= v(t)-1, divided by t-2

= [v(t)-1]/(t-2)

= (-32t+65-1]/(t-2)

= (-32t+64)/(t-2)

= -32(t-2)/(t-2)

=-32 ft/sec

but if you really meant initial velocity was 50 ft/sec, then recalculate the above

h(t) = -16t^2 +50t

v(t) = h'(t) = -32t +50

v(2) = -32(2)+50 =-64+50 =-14

average velocity

= (v(t)--14)/(t-2)

= (-32t +50)/(t-2)

=2(25-16t)/(t-2)