At the moment when a shotputter releases a 7.26 kg shot, the shot is 2.0m above the ground and travelling at 15.0 m/s. It reaches a maximum height of 8.0m above the ground and then falls to the ground. Assume that air resistance is negligible.
What is the final velocity of the shot? [VECTOR]
Solve with energy equations.
The mass is only included in the problem statement to confuse the student.
First do the vertical problem.
It had initial vertical velocity component Vi and height 2 meters
It went up to 8 meters and then stopped and fell.
How long to rise from 2 m to 8 m?
v = Vi - 9.81 t
at top, Vi = 0
so
t = Vi/9.81
Now the height part
8 = 2 + Vi t - (9.81/2) t^2
or
6 = Vi (Vi/9.81) - (9.81/2) (Vi/9.81)^2
6 = Vi^2 [ 1/9.81- (1/2) /9.81 ]
6 = Vi^2 (.5/9.81]
Vi = 10.8 m/s
It is easiest by considering energy.
final KE=initial KE + intial PE
1/2 m vf^2=1/2 m vi^2 + mgh
if you do some algebra, this will become
vf^2=vi^2 + 2gh
h=2, g=9.8 vi=15
solve for vf
Now do the second half of the vertical problem
I did not notice you said use energy so I will now
Call vertical component at ground Vf
fall from 8 meters
(1/2) m Vf^2 = m g (8)
Vf = sqrt (16 g)
Now you need horizontal speed.
That is easy
You have Vi = 10.8
and Vi^2 + u^2 = 15^2
so
u = sqrt (225 - 117)
and then finally
speed out of cannon = sqrt (Vi^ + u^2)
How can I find the vector if energy works in scalars? I can find the final speed without problem but I can't break it down into components or figure out how to get the direction at all
I gave you components
you have vertical component at ground = Vf
and find horizontal component the way I did
Watch out Bob, 15 m/s is not the vertical component but the total speed
If you want to do that first part with energy
we do NOT know Vi
BUT
we do know that the thing goes up 8-2 = 6 meters before gravity stops it
Vi = sqrt (2 g * 6 meters)