projectiles ... Been doing this problem for over an hour, PLEASE HELP will give 10 pts?

Question

I cannot figure out how to do this problem...

The projectile from a be-be-gun which is pointed vertically up reaches a maximum height h1. The maximum height when the same gun is tilted at an angle of 53.54 degrees with respect to the horizontal is h2. What is the ratio h2/h1 ?

i do not want the answer, that will not help me, please explain how to get the answer ... the correct answer for it btw is .6468533363118013 so u can use that the check ur method.


Please help

um, i tried pluggin in the numbers with that first answer and it didnt come out right... can u please explain more mayb, or check ur work

Answer 1

Ok, since we only care about height, let's only consider the VERTICAL components of the 2 situations.

now, distance travelled is still going to be our regular equation:
s = s0 + v0*t + 1/2 * g * t * t

In both cases, s0 = 0, and v0 = the vertical muzzle velocity, and g is a NEGATIVE if the velocity is positive, so....
s = 0 + v0*t - 1/2 * g * t * t

Now the time to apex (max height) is quite simply v0/g (the time it takes for the gravity to decelerate the bb), so....
s = v0*v0/g - 1/2 * g * v0/g * v0/g
s = v0*v0/g - 1/2 * v0*v0/g (note some of the g's cancel)
s = 1/2 * (v0 * v0) / g

so now we know the distance travelled upward is a direct relationship of the SQUARE of the muzzle velocity upward.

Now, in the first situation the muzzle velocity upward is the same as the muzzle velocity total (since the barrel was pointed up), but in the 2nd, the bb's upward velocity is the SIN of the muzzle velocity. Imagine a right triangle with the right angle at the lower right, a 53.54 degree angle at the lower left, and the other angle above the right angle. The gun is at the lower left. Since the gun is shooting along the hypotenuse, the vertical component is the SIN and the horizontal component is the COS.

So, since the SIN of 53.54 = 0.804272, that's the vertical component of the muzzle velocity for the 2nd situation. Since the height traveled is relational to the SQUARE of that velocity, we square the number to get 0.64685333... your solution.

Answer 2

This solution really just requires a knowlege of how to manipulate vectors.

So start by computing the maximum height of the bebe. The bebe will travel up until Vy (its velocity in the y-direction) equals zero:

0 = Vy - gt (where g is the downward acceleration due to gravity)

=> t = Vy/g : this is the time when the bebe has reached maximum height.

Plug this back into the formula for displacement:

y = Vy* t - g/2 * t^2

=> Vy (Vy/g) - [g(Vy/g)^2] /2

This simplifies to (Vy^2)/2g.

So then to get the ratio of h1 to h2 we just have to compare initial velocities in the y-direction. Assuming every bebe leaves the gun at the same speed, when the bebe if fired straight up, then 100% of the initial velocity is in the y-direction.

But when the gun is fired at an angle, only some of the initial velocity is in the y-direction... but how much? You can think of it as a triangle. The cosine of the angle that the firing angle forms with the ground is the x-component of the velocity, and the sine is the y-component. This means the ratios of the velocities are sin(53.54)/1

If you want a ratio of the heights, just plug in :

h2/ h1 = [(Vy_2)^2/g] / [(Vy_1)^2/g]

the g's cancel, leaving

h2/h1 = (Vy_2)^2 / (Vy_1)^2

Vy_1 = Vo

Vy_2 = sin(53.54) * Vo

this means the ratio of the squares is just [sin(53.54)]^2.

Plug this into your calculator & you'll find it's the correct answer as listed in your question.

Answer 3

In the vertical direction:
Here, you're looking for the parameters that describe a projectile that moves straight up with an initial velocity equal to the muzzle velocity (the velocity of the bb as it emerges from the rifle barrel) and influenced by gravity.. At the top of its flight, the velocity is zero.

Recall (Vfinal)^2 = (Vinitial)^2 + 2 * a * s
Here Vfinal = 0, Vinitial = Vmuzzle, a = g = -9.8 m/s^2, s = h1
Substituting,
0 = Vmuzzle^2 - 2*9.8 m/s^2 * h1
Vmuzzle = squareroot(19.6 * h1)

At an angle of 53.54 degrees:
Here, all you care about is what goes on in the vertical direction. If the velocity along the angle is the muzzle velocity, then the vertical component of the bb's velocity is given by
Vvert = Vmuzzle * sin(53.54 deg)
= squareroot(19.6 * h1) * sin(53.54 deg)
=squareroot(h1) * squareroot(19.6) * sin(53.54deg)
=squareroot(h1) * 3.56

Again, the equation to use is
(Vfinal)^2 = (Vinitial)^2 + 2 * a * s
where now Vinitial = Vvert, Vfinal = 0, a = g, s = h2
0 = [squareroot(h1) * 3.56]^2 + 2 * (-9.8 m/s^2) * h2
h1 * 12.678 = 19.6 * h2
12.678/19.6 = h2/h1 = .64685

Answer 4

the method is very simple

let the velocity with which the bullet comes from the gun =u.
at the max height the vel is 0.
the max height (according to the problem) is h1.
the max height then can be found by the formula v^2=u^2-2gh1
where v=0.
hence
u^2=2gh1............(1)
for firing at an angle we consider the vertical component of the velocity.
the vertical component can be obtained by resolving the forces one along the vertical and one along the horizontal.
the vertical component is given by u*sina where a is the angle of projection(=53.54 degrees)
hence here we have (usina)^2=2gh2.......(2)
(2)/(1) gives
(sina)^2=h2/h1
=>(sin53.54)^2=h2/h1
therefore h2/h1=0.6468533363118013

Answer 5

Try the site below. This is the ratio of the effect of the acceleration of gravity factor considered vertically (h1) and at the specified angle (h2). The mass and energy must also be known for practical computations.

http://www.morpheus.cc/ITESM_phys/default.htm

Answer 6

Introduction


Discus, javelin, shot, and hammer throwing are Olympic field events involving projectiles. Other Olympic events involve jumping. For example, while competitors in the long jump frequently hurl themselves over 7 m into a sandpit, ski jumpers can travel over 100 m between take-off and landing.

Answer 7

Please tell me she's a minimum of paying you for this, if no longer right it is my suggestion: tell your Dad your uncomfortable with a concern that is composed of your so called abode college instructor. Ask for a gathering. tell her the way you experience together as your Dad is latest- be user-friendly and open. enable her be attentive to you experience as though she's taken your kindness with no attention. And tell her, confident you're a to blame youthful lady yet, you may in basic terms cope with plenty at your age. additionally, her activity is barely to show and that i'm useful the challenge of "Parenting" is slightly too early so which you will digest. do no longer hesitate to experience as though your disrepecting your elders by using fact your in basic terms status up for your self. you sound like a great individual who does not should be a pushover. i'm hoping each little thing works out for you! stable success newborn!

Answer 8

h=v^2sin^2t/2g if it is projected at angle t with horizontal

hence h1/h2=sin^2(t1)/sin^2(t2)

put the value t1=90 t2=53

Answer 9

The second shot only gives a portion of the initial velocity upward, Sin(53.54) to be exact.

The alttude that you reach is porpotional to you inti energy, which is the square of your init vel, (sin(53.54)^2 = 0.646....