# Falling Bodies?

• May 12th 2008, 06:03 PM
keemariee
Falling Bodies?
If a ball is thrown vertically upward from the roof of a 64 foot building with a velocity of 80 ft/s, it's height after t sexonds is s(t)=64+80t-16t^2. What is the maximum height the ball reaches? What is the velocity of the ball when it hits the ground?
A body is thrown straight up with initial velocity 5 feet per second from a height of 40 feet. After how many seconds will it hit the ground? What will be its maximum height?
• May 12th 2008, 06:26 PM
Chris L T521
Quote:

Originally Posted by keemariee
If a ball is thrown vertically upward from the roof of a 64 foot building with a velocity of 80 ft/s, it's height after t sexonds is s(t)=64+80t-16t^2. What is the maximum height the ball reaches? What is the velocity of the ball when it hits the ground?
A body is thrown straight up with initial velocity 5 feet per second from a height of 40 feet. After how many seconds will it hit the ground? What will be its maximum height?

You can use calculus or algebra to solve this. Do you know calculus or algebra or both? I don't want to do a problem one way when it should be done another way!!
• May 12th 2008, 07:47 PM
keemariee
i'm finishing up in a calculus class... its basically self-taught so it's extremely hard and we're all struggling to understand everything fully.
• May 12th 2008, 08:09 PM
Chris L T521
Quote:

Originally Posted by keemariee
If a ball is thrown vertically upward from the roof of a 64 foot building with a velocity of 80 ft/s, it's height after t sexonds is s(t)=64+80t-16t^2. What is the maximum height the ball reaches? What is the velocity of the ball when it hits the ground?
A body is thrown straight up with initial velocity 5 feet per second from a height of 40 feet. After how many seconds will it hit the ground? What will be its maximum height?

Quote:

Originally Posted by keemariee
i'm finishing up in a calculus class... its basically self-taught so it's extremely hard and we're all struggling to understand everything fully.

since we have our position function $\displaystyle s(t)=64+80t-16t^2$, we can differentiate it to find the velocity:

$\displaystyle v(t)=s^{/}(t)=80-32t$. The projectile is at it's maximum height when $\displaystyle v(t)=0$. Thus, we see that:

$\displaystyle 80-32t=0 \rightarrow 32t=80 \rightarrow t=2.5$ seconds. To find the value of the maximum height, we plug the t value into the position function, s(t).

max height: $\displaystyle s(2.5)=64+80(2.5)-16(2.5)^2=\color{red}164\space ft$.

to find the maximum velocity when it strikes the ground, find where $\displaystyle s(t)=0$ and then plug the value into v(t).

$\displaystyle 64+80t-16t^2=0 \rightarrow t^2-5t-4=0$ when $\displaystyle t=\frac{5\pm \sqrt{25+16}}{2}=-.702\space or\space 5.702$.

We take the positive value and substitute it into velocity:

$\displaystyle v(5.702)=80-32(5.702)=-102.45$. The negative tells us its direction (down). It will have a velocity of $\displaystyle \color{red}102.45\frac{ft}{s}$ as it hits the ground.

For the second question, first come up with s(t).

$\displaystyle s(t)=s_0+v_0 t+\frac{1}{2}gt^2$. We know $\displaystyle s_0=40\space ft$, $\displaystyle v_0=5\frac{ft}{s}$, and $\displaystyle g=32\frac{ft}{s^2}$. Our equation now becomes:

$\displaystyle s(t)=40+5t-16t^2$.

Why don't you try this one yourself? If you get stuck, I'll be here to help.