Sunday

November 23, 2014

November 23, 2014

Posted by **Marie** on Friday, April 20, 2012 at 4:41pm.

Tower with an initial velocity of – 20 feet per second (the negative velocity indicates that the object

is moving towards the ground). Given that the acceleration due to gravity of the object is a constant –

32 feet per second per second, determine how far above the ground the object is exactly six seconds

after being thrown.

- Applied Calculus -
**Damon**, Friday, April 20, 2012 at 6:59pmh = Hi + Vi t + (1/2) a t^2

Hi = 1100

Vi = - 20

a = -32

so

h = 1100 -20 t - 16 t^2

if t=6

h = 1100 -20*6 - 16*36

= 404 ft

- Applied Calculus -
**Damon**, Friday, April 20, 2012 at 7:02pmall this from integrating acceleration

a = -32 ft/s^2

so

dv/dt = -32

v = Vi - 32 t

dh/dt = v = Vi - 32 t

h = Hi + Vi t - (1/2)(32)t^2

**Answer this Question**

**Related Questions**

Algebra - THe function h(t)=-16t^2+v0t+h0 describes the height in feet above the...

Algebra - THe function h(t)=-16t^2+v0t+h0 describes the height in feet above the...

College Algebre - Here is the math problem: The height of an object in free fall...

calculus - use a(t) = -32 ft per second squared as the acceleration due to ...

Calculus - When a ball is thrown straight down from the top of a tall building, ...

12th Grade Calculus - When a ball is thrown straight down from the top of a tall...

algebra - y=(v/u)x+(g)/(2(u^2))*x^2 where y is the object's vertical distance (...

math - an object is thrown upward from the top of a 80-foot building with a ...

Math - An object is thrown from the top of an 80 foot building with an initial ...

Math - The height(h) of an object that has been dropped or thrown in the air is...