Projectile Motion (1 Viewer)

[lyounamu]

A shell is fired at 200 metres/second to strike a target 2 km away. At what angle should it be fired?

This question looks so simple and easy but it is a bit weird. What I did was:
V = 0.2 and the coordinate is (2,0)

I used the equation of the path to find the angle but I failed.

Part of my working out is like the following:
y = x tanx - gx^2/2V^2 (1+tan^2(@))
Substitute x = 2, y = 0 and V = 0.2
0 = 2 tan x - (10 . 2^2)/(2 . 0.2^2) x (1+ tan^2(@))
= 2 tan x - 500 (1+tan^2(@))
= 2tan x -500 - 500 tan^2(@)
And I cannot seem to work this out.

 

[ronnknee]

Mm let me see

Edit:
I would say you have taken the wrong approach. Instead of using the equation of the path and solving for theta, why don't you consider making use of the range?

y = V sin @ t - 1/2 gt²
x = V cos @

When y = 0, V = 200, g = 10
0 = 200 sin @ t - 5 t²
0 = 5t(40 sin @ - t)
Therefore t = 40 sin @

When t = 40 sin @, x = 2000, V = 200
2000 = 200 cos @ . 40 sin @
10 = 40 sin @ . cos @
1/4 = sin @ . cos @
1/2 = 2 sin @ . cos @
sin 2@ = 1/2
2@ = 30 degrees or 150 degrees
@ = 15 degrees or 75 degrees


Edit 2:
A flaw in your approach is that you're taking the units of displacement as kilometres and acceleration of gravity as metres. The inconsistency may have contributed to the undoable quadratic?

 

[u-borat]

...solve it as a quadratic maybe?

 

[lyounamu]

...solve it as a quadratic maybe?

I did. It's a positive definite.

 

[ronnknee]

Read my second edit

 

[lyounamu]

Mm let me see

Edit:
I would say you have taken the wrong approach. Instead of using the equation of the path and solving for theta, why don't you consider making use of the range?

y = V sin @ t - 1/2 gt²
x = V cos @

When y = 0, V = 200, g = 10
0 = 200 sin @ t - 5 t²
0 = 5t(40 sin @ - t)
Therefore t = 40 sin @

When t = 40 sin @, x = 2000, V = 200
2000 = 200 cos @ . 40 sin @
10 = 40 sin @ . cos @
1/4 = sin @ . cos @
1/2 = 2 sin @ . cos @
sin 2@ = 1/2
2@ = 30 degrees or 150 degrees
@ = 15 degrees or 75 degrees


Edit 2:
A flaw in your approach is that you're taking the units of displacement as kilometres and acceleration of gravity as metres. The inconsistency may have contributed to the undoable quadratic?

I see. Thanks!

 

[u-borat]

ronnknee, his method was pretty much the same as yours; you solved for time using one motion equation then subbed it into the other motion equation.
the method i prefer, and the one namu prefers is the elimination of the time...and yeah, good pickup on the wrong use of units.

 

[lyounamu]

ronnknee, his method was pretty much the same as yours; you solved for time using one motion equation then subbed it into the other motion equation.
the method i prefer, and the one namu prefers is the elimination of the time...and yeah, good pickup on the wrong use of units.

Yeah. I cleverly changed the unit for the veloctiy but I forgot the gravity part. How dumb...

I am using Fitzpatrick at the moment and until Q10, (which was the question I asked), there was no question relating to the change in unit. That's why I probably screwed it up. Not bad for the newbie who started Projectile today, eh?

 

[vds700]

Yeah. I cleverly changed the unit for the veloctiy but I forgot the gravity part. How dumb...

I am using Fitzpatrick at the moment and until Q10, (which was the question I asked), there was no question relating to the change in unit. That's why I probably screwed it up. Not bad for the newbie who started Projectile today, eh?

theres an evil question in there i think its 17, about throwing a stone at a bird lol

See if u can get it

 

[lyounamu]

theres an evil question in there i think its 17, about throwing a stone at a bird lol

See if u can get it

LOL

As if I can get it. The image of a stone hitting a bird further deteriorated my thinking ability. I think I got my drawing right but there are two velocity involved.

In conclusion, I cannot do it. If you know the solution, please post it up so that I can see. Thanks!

EDIT: I will post this up just to let everyone know.

A stone is thrown so that it will hit a toy at the top of a pole. However, at the instant the stone is thrown, the toy flies away in a horizontal straight line at a speed of 10m/s. The stone reaches a heigh double that of the pole and, in its descent, touches the tory. Find the horizontal component of the velocity of the stone. (I changed bird to a toy, making it less evil. You don't have to image a terrible scene this way).

 

[ronnknee]

ronnknee, his method was pretty much the same as yours; you solved for time using one motion equation then subbed it into the other motion equation.
the method i prefer, and the one namu prefers is the elimination of the time...and yeah, good pickup on the wrong use of units.

I initially thought his approach was wrong because he ended up with a undoable quadratic. So instead, I took another approach and yielded the correction. After looking at his approach the second time, I realised that there was an inconsistency with the units which led to the undoable quadratic. So yes I acknowledged that his approach was correct but was flawed with the inconsistency of units

 

[ronnknee]

http://community.boredofstudies.org/showthread.php?t=176770

For the post on the bird and stone

I couldn't do the question either

 

[vds700]

A stone is thrown so that it will hit a toy at the top of a pole. However, at the instant the stone is thrown, the toy flies away in a horizontal straight line at a speed of 10m/s. The stone reaches a heigh double that of the pole and, in its descent, touches the tory. Find the horizontal component of the velocity of the stone. (I changed bird to a toy, making it less evil. You don't have to image a terrible scene this way).

you should have said a pigeon instead

 

[lyounamu]

you should have said a pigeon instead

pigeon = bird in my eyes.

I didn't want to hurt those poor creatures.

UPDATE: UP to Q 14 now. Got everything so far (except the question I asked).

 

[vds700]

I didn't want to hurt those poor creatures.

lol