How to Calculate Projectile Angle for Landing in a Specific Distance - 120m

[bengaltiger14]

Help with a Problem Please...

This is the problem.

A projectile is being shot from a cannon and must land in a hole 120 meter aways. What angle must the projectile be shot at to land in the hole? Here are some other numbers


VoX=60cos(??) Voy=60sin(?)
VfX=? Vfy=?
Ax=0 Ay=-9.81 m/s^2
t=?? t=??
dx=120m dy=0

 

[bengaltiger14]

Sorry, just read the post that I need to post homework problems elsewhere.

 

[Delzac]

take note that the time for vertical and horizontal motion is the same.

since you know the horrizontal velocity, which is in terms of 60cosZ, can you find time?

 

[Hootenanny]

One is also expected to show ones efforts.

 

[bengaltiger14]

I tried finding t this way dx=Vxt === 120m = 60cosZt

t=120/60cosZ

 

[Delzac]

yeah you are on the right track, try equate the time into your vertical motion equation.

( they land on the ground the same time, the 2 motions)

 

[bengaltiger14]

The answer would be 0 as far as the vertical aspect is concerned right?? I mean the vertical displacement is 0.

 

[Delzac]

correct vertical displacement would be zero

EDITED version

 

[bengaltiger14]

So, would the vertical time be t = sqrt d/VoYt + 1/2a

 

[Delzac]

you can simplify the equation into 0 = t(-Voy + 1/2 g t) ( when solving this type of qns remember to state your sign convention, you don't want the case where you get negative time)

so Voy = 1/2 g t

 

[bengaltiger14]

So t=VoY/2g. But how do I find that if I do not know the angle?? You can't divide 60sin of nothing by 2g can you?

 

[Delzac]

you can equate the 2 equations u have together, the horizontal and vertical, remember the time is the same for both.

 

[bengaltiger14]

Equate them together? So... t= (120/60cosZ) * (60sinZ/2g)

 

[Delzac]

if you equate in such a way, you have 2 viarables to solve and only one equation to work with.

BTW your equation should be T^2 = ..., since u times the 2 equation together.( but this is not the correct equation to use.)

 

[bengaltiger14]

So that is not the correct way to equate it?? I am really confused here.

 

[Delzac]

T = 120/ (60 cos Z)

T = 2(60 sin Z)/(9.81)

120/(60 cos Z) = 2(60 sin Z)/(9.81)

EDIT : i amended the equations. For the time derieved by vertical motion,

Voy = 1/2 g t
(2 voy)/g = t

t = (2 (60) sin Z) / g

You wrote wrongly the equation for your previous post.

 

[bengaltiger14]

Ok. I'm sorry. I thought I put a equals between the two equations. I did not realize it was a muliplication sign. What do I solve for now?? Do I set the equation equal to 0 and divide -120 by 2(9.81)?

 

[Delzac]

you can cross mutiply. Then isolate your Z, i used double angle formula for this.

BTW this is the first time i use double angle formula for physic. feels strange

 

[bengaltiger14]

So, 120 m * (2*9.81) = 2354s^2. What would 60 sinZ time 60cosZ be? 3600 sinZ?

 

[Delzac]

So, 120 m * (2*9.81) = 2354s^2.

this part is wrong, as i have stated about, the equation is

120/(60cos Z) = ((2)(60)sin Z)/ 9.81

thus the correct expression should be 120(9.81) = (2)(60)(60)cos Z sin Z

 

[Delzac]

Then use double angle formula :

\sin(2x) = 2 \sin (x) \cos(x) \ = \frac{2 \tan (x)} {1 + \tan^2(x)}

 

[Delzac]

Z should be 9.54339 degree

 

[bengaltiger14]

Ok. I am not familiar with double angle forumula. So far I have gotten.

I multiplied and got 1177s^2 = 7200 cosZSinZ

 

[bengaltiger14]

Ok. I see the forumla now. Thanks

 

[Delzac]

now the doucle angle formula states that:

\sin(2x) = 2 \sin (x) \cos(x)

so we shall express our trigo in this form:

(1177.2)/( 60*60) = 2 sin Z cos Z

2 sin Z cos Z = sin (2Z) as stated by the formula.

There solve from here : sin (2 Z ) = (1177.2)/(60*60)

 

[bengaltiger14]

so sin(2 Z) = .327 Do I divide that by 2. What do I do with the 2Z??

 

[Delzac]

2Z = inverse sin ( 0.327)
2Z = 19.08678...

angle is equal to Z.

 

[bengaltiger14]

Thank you for all your help. As, you can tell, Trig is my weak spot and I need to focus hard on that I see now.

 

[Delzac]

Welcome, always a pleasure to help other.

BTW welcome to PF

 

[bengaltiger14]

Thank you. I am thankful there are people like you around to help others who need help.