# Differential Equations(Related Rates)

• Sep 26th 2008, 05:43 PM
Oblivionwarrior
Differential Equations(Related Rates)
I am stuck on this question, here it is.

A ball with mass m kg is thrown upward with initial velocity 28m/sec from the roof of a building 11m high. Neglect air resistance.
(a) Find the maximum height above the ground that the ball reaches.
(b) Assuming that the ball misses the building on the way down find the time that it hits the ground.

I found the answer to part (a) as 51 meters but I can't get part (b). Thanks for any help!
• Sep 26th 2008, 06:12 PM
Chris L T521
Quote:

Originally Posted by Oblivionwarrior
I am stuck on this question, here it is.

A ball with mass http://edugen.wiley.com/edugen/share...=1222472761595 kg is thrown upward with initial velocity http://edugen.wiley.com/edugen/share...=1222453812839 m/sec from the roof of a building http://edugen.wiley.com/edugen/share...=1222453812839 m high. Neglect air resistance.
(a) Find the maximum height above the ground that the ball reaches.
(b) Assuming that the ball misses the building on the way down find the time that it hits the ground.

I found the answer to part (a) as 51 meters but I can't get part (b). Thanks for any help!

I don't see the numerical value for speed and for height. Is the mass assumed to be some value $m$??

--Chris
• Sep 26th 2008, 06:22 PM
Oblivionwarrior
Quote:

Originally Posted by Chris L T521
I don't see the numerical value for speed and for height. Is the mass assumed to be some value $m$??

--Chris

Thanks! fixed.
• Sep 26th 2008, 06:31 PM
skeeter
$h(t) = 11 + 28t - 4.9t^2$

part (b) ... set h(t) = 0 and solve for t.
• Sep 26th 2008, 06:38 PM
Chris L T521
Quote:

Originally Posted by Oblivionwarrior
I am stuck on this question, here it is.

A ball with mass m kg is thrown upward with initial velocity 28m/sec from the roof of a building 11m high. Neglect air resistance.
(a) Find the maximum height above the ground that the ball reaches.
(b) Assuming that the ball misses the building on the way down find the time that it hits the ground.

I found the answer to part (a) as 51 meters but I can't get part (b). Thanks for any help!

I broke up part (b) into 2 parts:

from 11 m to 51 m (from top of building to highest point):

Using $y_f=y_0+v_0t-\tfrac{1}{2}gt^2$, we see that $51=11+28t-4.9t^2\implies 4.9t^2-28t+40=0$

I'll let you solve this and show that $t\approx 2.86~s$

from 51 m to 0 m (from highest point to the ground):

Using $y_f=y_0+v_0t-\tfrac{1}{2}gt^2$, we see that $0=51-4.9t^2\implies 4.9t^2=51\implies t^2=10.41\implies t\approx 3.23~s$

Thus the total time for it to reach the ground, once it was thrown off the building is $t=2.86~s+3.23~s=\color{red}\boxed{6.09~s}$

Does this make sense?

--Chris
• Sep 26th 2008, 06:39 PM
Chris L T521
Quote:

Originally Posted by skeeter
$h(t) = 11 + 28t - 4.9t^2$

part (b) ... set h(t) = 0 and solve for t.

haha...wow...I can't believe I didn't think of that... >.<

--Chris