Velocity Problem

[sb_4000]

Hi,
Im haveinga problem setting up an equation.

1) A ball is thrown upward with initial velocity of 80 m/s.
a) how long is the ball in the air?
b) what is the greatest height reached by the ball?
c) when is the ball 20 m above the ground?
d)what is the velocity of the ball, when it hits the ground?

Im having problem setting up the equation in part d.

Thank You.

[NateTG]

There's a clever way to find the answer, and a not so clever one that involves using the time that the ball is in the air (wich you got as part a).

[sb_4000]

for part a

v=v-0 +at
t= 2(v_0-V)/g = 2(8.16) = 16.33s

this is how I got part a.

[NateTG]

So, what's wrong with using:
v=v_0+at
for part d?

P.S. The clever answer is to recognize that the situation is symetric.

[sb_4000]

i get -80 is this correct for part d?

[NateTG]

i get -80 is this correct for part d?
Yes. (That is correct, this text is there to satisfy the rather silly board software.)

[sb_4000]

thanks alot Nate..
I have one question, for projectile motion would I be able to use the same method for solving the velocity of the ball when it hits the ground.

Here is the question.

Projectile is fired from the top of a 40m tower at an angle of 60 deg above the horizontal hits the ground at the point 100m from base of the tower.

a)find the speed at which the stone was thrown- I got 35 m/s
b) find the speed of the stone just before it hits the ground?
c) find the time at which it hits the ground?
d) find the height- I get 46.88 m

can you check and see if the answers are right..I wasent able to get part b and c..but i did some work and got -41.132 m/s for B, and 7.29 s for C..

Thank You.

[NateTG]

Ok, that's much tougher:

\vec{v_0}=< v_{0x},v_{0y} >
From 60 degrees we get:
\tan(60)v_{0x}=v_{0y}
\sqrt{3}v_{0x}=v_{0y}

I'm going to set up the ground as zero.
Then:
x(t)=v_{0x}t
and
y(t)=40+v_{0y}t - \frac{1}{2} gt^2

so let's make some substitutions:
x(t)=\frac{1}{\sqrt{3}} v_{0y} t
since we want to know the time for 100 meters out:
100=\frac{1}{\sqrt{3}} v_{0y} t \rightarrow v_{0y}=\frac{100\sqrt{3}}{ t}
and the final height is 0
0=40 + v_{0y}t - \frac{1}{2} gt^2
substitute for v_{0y}
0=40 + 100\sqrt{3} + - \frac{1}{2} g t^2
t = \sqrt {\frac{80 + 200\sqrt{3}}{g}}
t \approx 7.1s
v_{0y} \approx 24 \frac{m}{s}
v_0 \approx 28

So, I get an inital velocity of 28m/s, and a time of 7.1 seconds.
Now,
v_{x,final}=v_{x0}\approx 14\frac{m/s}
and
v_{y,final}=v_{y0}-gt \approx 24-70 = -46
so
v_{final} \approx \sqrt{14^2 + 46^2} \approx 48

To find the maximum height, set v_y=0 and solve. I get 68m.

Of course, I may have made some errors in my work.

[sb_4000]

Thanks Nate, I appreciate the help.

[sb_4000]

for t i get 6.59

[NateTG]

for t i get 6.59

Well, you can plug your results back into the equations of motion, and see if they come out right to check your answers.