Calculus
posted by Jacke on .
1.if a ball is thrown into the air with a velocity of 35ft/s, it's height in feet after t seconds is given by y=35t10t^2. Find the averag velocity for the time period beginning when t=4 and lasting 0.05s.
2.if a ball is thrown into the air with a velocity of 30ft/s, it's height in feet after t seconds is given by y=30t20t^2. Find the instantaneous velocity when t=3

Last time I was on this earth, gravity would cause an equation of the type
y = 16t^2 .....
So assuming you have a typo and it should have been
y = 35t  16t^2
Second problem:
the ball would not be in the air for 4 seconds,
so your whole first question is actually just gibberish.
Once you fix it, find y when t = 4
find y when t = 4.05
evaluate: ( y(4.05)  y(4) )/.05
2. same thing here,
after 3 seconds the ball has reached the ground, and no longer has any velocity
if we ignore reality,
dy/dt = 30  40t
when t = 3
dy/dt = 30  120 = 90 ft/s
1.
when t=4, y = 35(4)  16(16) = 20 
There is no error in the problem.

I see the typo as specifying feet. In metric units, g=9.8, or roughly 10, so Jacke, redo the calculations, with g=10 and get the correct answer.

Is it ok if I get a negative answer???

OK thanks I got it