in this video we have a sort of cumulative review kind of problem i have a bowling ball with a mass
of five kilograms it's being launched by a spring with constant 120 newtons per meter the initial
compression is 2.5 meters and the angle of this spring gun is given to us and so the first part
of the problem is to figure out if i launch this thing how fast is the ball going when
it flies out of the gun and i'm going to refer to that as v naught and then the second part of
the problem is well now that i know how fast it's launched out of the tip of this giant spring gun
how far is it going to go before it hits the ground and that's a projectile motion problem
so the first part is approached with energy conservation i have energy stored in the spring
but here's a warning in part a you can't just ignore gravitational potential energy and pretend
that the spring energy all goes into the speed of the ball because the ball ends up higher than
it was in its initial position so that spring energy goes to two places part of it is the
increase in gravitational potential energy part of it is the increase in kinetic energy and i'm
going to be clear about my coordinate system here for the first part i'm going to call y equals zero
at the initial position of the ball then the final position of the ball when it pops out of this gun
that's going to be y final and i need to find out how far up did it go during this process because
that's going to tell me how much of the energy went into the gravitational potential energy
and that's just given by a leg of a right triangle where 2.5 meters the compression distance that's
the hypotenuse the angle here is 40 degrees and 2.5 sine of 40 will give me that vertical change
and i'll go ahead and pop that right into the diagram to three sig figs that's 1.61 meters of
increase in height during the launching process so we write down our statement of energy conservation
e initial is equal to e final during the launch process and the initial energy because i put the
origin of coordinates at the initial location of the ball the y-coordinate is zero so the
gravitational potential energy is zero and there's certainly no kinetic energy in the initial state
so the only energy in the initial state is the potential energy stored in the spring and then
in the final state when the ball pops out of the gun i have some gravitational potential energy
and i have some kinetic energy so we plug in the expressions for all these things and my initial
spring potential energy is one half kx squared i'll call it x initial for the initial compression
my final gravitational potential energy is going to be an mgy final and my final kinetic energy is
going to be a one-half m v squared and i'm going to call that v naught because that's the launch
velocity we're going to use for part b so we just plug in all the numbers k was 120 newtons per
meter the initial compression was 2.5 meters the mass of the bowling ball is 5 kilograms g is 9.8
my final y coordinate when the launch is complete that's 1.61 meters and again m is 5 kilograms and
my only unknown there is the speed at the end of the launch process so i prefer to do calculations
like this just all at once in my calculator so i'm going to smash these numbers together on the left
and then i'm going to subtract the product of all these numbers then multiply the result by 2 divide
by 5 and square root the result and we've got v naught and i get a launch velocity of 10.88 meters
per second into three sig figs that's 10.9 meters per second okay so that's part a and then in part
b i want to compute the range of the ball and this is just a classic projectile motion problem
i have a launch velocity that's known i'll go ahead and call it 10.88 meters per second
so i don't compound the rounding error more than necessary i know the initial launch height that's
given in the diagram that's an initial height of 2 meters and i want to get the range how far does
it go horizontally before it hits the ground and so the standard approach to this is to decompose
that initial velocity vector into components and the horizontal one i'll call v naught x
and just a reminder this angle here is still 40 degrees so v naught x is 10.88 cosine of 40
and that's about 8.33 meters per second then v naught y is going to be a 10.88 sine of 40. and
this gives me about 6.99 meters per second then what we know in the final state of the projectile
so it hits the ground right here what we know is the final y-coordinate the final y-coordinate is
zero so we're going to use a y analysis first and we're going to use that to find the flight time
so we write down our first constant acceleration kinematics formula the vertical acceleration here
is negative g so i'll go ahead and plug that in and the final y position is zero the initial y
position was two meters initial y velocity was 6.99 i don't know the flight time half of g is
4.9 if we use 9.8 for g so i have a minus 4.9 t squared on there and we have a non-trivial
quadratic equation to solve so we're going to use the quadratic formula for this i prefer to have a
positive leading coefficient on that so i'm going to add all the terms to the left side equivalently
you could say i'm just multiplying all this by negative one and you can either plug this into
a piece of technology or we can go ahead and apply the quadratic formula i'll go ahead and show the
long way so t is going to be negative b plus or minus the square root b squared minus 4 times a
times c and there's a double negative in there to be careful with
and then divided by 2a which is 9.8 and when i simplify the square root piece i get a 9.38
for that and i can immediately tell i'm going to get one positive value of t and one negative value
and we discard the negative value so the negative value will happen from subtracting the 9.38
i'm only going to keep the one where i add 9.38 and this gives me a flight time of 1.67
seconds next we go to the equation that describes the horizontal position as a function of time
so x is equal to x naught plus v naught x t and in general you'd have a plus one half x acceleration
times t squared but we know for projectile motion the x acceleration vanishes this also means the
initial x velocity is just equal to the final x velocity it's just a constant velocity in the
x direction by assumption here we're going to put our initial position at zero and then the
final position will be the range so we plug in our initial x velocity and that was 8.33
meters per second we multiply it by how long it was moving that fast before it hit the
ground that was 1.67 seconds and it turns out the range of this bowling ball is 13.9
meters and we're done if you enjoyed this video or at least found it useful
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