# Calculus

posted by .

1. The problem statement, all variables and given/known data
a ball is thrown straight down from the top of a 220 feet building with an initital velocity of -22ft/s. What is the velocity of the ball after 3 seconds? What is the velocity of the ball after falling 108 feet?

2. Relevant equations
s(t) = -16t^2 + Vi(t) + si vi = initial velocity, si = initial position

3. The attempt at a solution
s(t) = -16t^2 +Vi(t) +si
= -16t^2 - 22ft/s + 200ft

V(t) = -32(t) - 22ft/s
V(3) = -96 ft/s - 22ft/s = -118 ft/s

I can seem to figure out how to determine the velocity of the ball after it has dropped 108ft.

I should be able to do this, but i'm missing something, somewhere on what should be pretty basic.

Thanks for any guidance!

• Calculus -

s(t) = 16t^2 - 22t + 220 is correct and tells you
how high the object is at any time t
But they are asking for the velocity after it has fallen 108 feet, not when it is 108 feet high

It started at 220, so after falling 108 feet, the height will be 112

so 112 = -16t^2 - 22t + 220
16t^2 + 22t - 108 = 0
t = (-22 ± √7396)/32
= 2 or a negative, (ahhh, it would have factored)

so take v(2) , that should give you the answer you need

check:
when t=0 , s(0) = 220 -- it has fallen 0 feet
when t=1 , s(1) = 182 -- it has fallen 38 ft
when t=2 , s(2) = 112 -- it has fallen 108 ft

## Similar Questions

1. ### Calculus

When a ball is thrown straight down from the top of a tall building, with initial velocity 30 ft/sec, the distance from the release point at time t in seconds is given by s(t)=16t^2+30t. If the release point is 300 ft above ground, …

When a ball is thrown straight down from the top of a tall building, with initial velocity 30 ft/sec, the distance from the release point at time t in seconds is given by s(t)=16t^2+30t. If the release point is 300 ft above ground, …
3. ### calculus

If a ball is thrown straight up into the air with an initial velocity of 70 ft/s, it height in feet after t second is given by y = 70 t - 16 t^2. Find the average velocity for the time period begining when t = 1 and lasting (i) 0.1 …
4. ### Calculus

A ball is thrown straight down from the top of a 220-foot building with an initial velocity of -22 feet per second. What is its velocity after 3 seconds?
5. ### math

A person standing on the roof of a building throws a ball directly upward. The ball misses the rooftop on its way down and eventually strikes the ground. The function s(t) = −16t2 + 64t + 80 describes the ball’s height above …

1. The problem statement, all variables and given/known data a ball is thrown straight down from the top of a 220 feet building with an initital velocity of -22ft/s. What is the velocity of the ball after 3 seconds?
7. ### Physical Science

Four balls are taken to the top of a building. Ball 1 is dropped (zero initial velocity). Ball 2 is thrown straight up with a velocity of 3 m/s. Ball 3 is thrown straight out with a horizontal velocity of 3 m/s. Ball 4 is thrown straight …
8. ### Calculus

If a ball is thrown straight up into the air with an initial velocity of 95 ft/s, its height in feet after t seconds is given by f(t)=95t−16t^2 Find the average velocity for the time period beginning when t=1 and lasting (i) …
9. ### Physics

Ball A is thrown straight down off the top of a tall building with an initial velocity of 20m/s. Ball b is thrown straight up from the top of the same building with initial velocity of 20 m/s. Which ball, if either, experiences the …
10. ### Math

A ball is thrown downward from the top of a 100-foot building with an initial velocity of 14 feet per second. The height of the ball h after t seconds is given by the equation h=-16t^2-14t+100. How long after the ball is thrown will …

More Similar Questions