[mrinformant]

Hey guys I have 3 different questions

1. A rifle aimed at an angle of 15deg from the horizontal was in the air 18,5s before falling back to the ground. What was its original speed?


2. The muzzle velocity of a mortar bomb is 100m/s. At what elevation should the barrel be pointed to achieve a range of 500m?


3. A projectile launched at 250m/s. At what angle must it be launched in order to rise to a height of 1880m?

Please reply asap
Cheers

[Curlyben]

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[Unknown008]

1. Call the initial velocity 'v'.

The vertical component is given by 'vsin15', right?

Now, using the equation

You can find the value of u, the initial vertical component, vertical displacement being 0, time being given and acceleration due to gravity being 9.8m/s^2. Find u, which is equal to vsin15, and solve for v.

Do that first and post what you have attempted for the other two problems to see where is your problem.

[mrinformant]

First of all, this isn't my homework
these are practice questions we were given to do at our own convenience
for question 1 my problem is: whilst I'm given time and degree I don't seem to have enough info to place into an equation
thanks jerry for your help but I'm still in the same situation
you said the vertical component is zero, whys that?

for question 2 I'm not sure what a muzzle velocity is and how it has an effect on the bomb + (Limited information, how do I approach question)

for question 3 I've got v = 250m/s and maxh = to 1880 but am unsure how to solve

[Unknown008]

Ok, I'll take the first one first.

You don't know the velocity of the projectile. You can have the vertical component of the velocity, in terms of v, which I called the velocity of the projectile.

Since you know the time, you can only consider the vertical displacements. If you had the distance the projectile covered from two points on the ground, then you'd need the horizontal displacement.

Let's take it like that. Say you have the projectile fired in front of you. You would only see it go up and then down, not any other direction to the right or left. Well, you know there is an initial vertical velocity (vertical component of the projectile). You don't know the horizontal one since you see it only go up then down.

What is the vertical component of the velocity?

You can find it using the formula I gave you above. I never said that vertical component is zero. I said the vertical displacement is zero. When you look at it from the front, you see it go up then down. Displacement = Distance between final and initial point. The projectile started on the ground, then went up, and ended back on the ground. No matter how high it went, it's displacement along the vertical axis is zero.

You know the time and the acceleration due to gravity. Use the equation to find the vertical component u. But u = v sin15. You now know u, and sin15, solve for v.

Post your answer.

[mrinformant]

I ended up getting it but I did it slightly different

I used v=u + at
v-u =at
v=-u therefore
-u-u = at
-2u = at
-2u = - 9.8 x 18.5
-2u = -181.3
u = 90.65m/s
v = 90.65/sin15

I'm really having trouble with question 2 though
any help would be great

[Unknown008]

Yes, that's also right. And you got the velocity v I hope? 350 m/s ?

Range depends on horizontal velocity.

The horizontal component is where theta is your angle of elevation.

Along the horizontal, you don't have acceleration, so, used speed-distance-time formula, s = ut.

Along the vertical, you have the vertical component as

The time is given from v = u+at. Since v = -u, you have -2u/a = t.

Bringing both equations, you have:



Sub in your values:



Solve for theta.

If you have a good memory, you can make use of the formula for range, but it cannot be used in all circumstances. It cannot be used if your projectile ends up on another height from where it was thrown.

The formula is

[mrinformant]

I've followed it but I think I've made an error somewhere ill show my working.

& will be my symbol for theta

500/100cos(&) = -2(100sin(&))/-9.8

4900 = -2(100sin(&))(100cos(&))
2450 = (100sin(&)) (100cos(&))
2450 = 10000 sin(&) cos(&)
0.245 = sin(&)cos(&)
that's where I'm up to

if I do individually sin(&) = 0.245 and likewise for cos I get the right answer but I don't think I can just do that. Have I made any errors or is there anything I haven't done?
thanks

[Unknown008]

No, you cannot do that. You cannot say sin(theta) = 0.245 or cos(theta) = 0.245.

You need a trigonometric proof here, which is:



With that, continue your work.