Help with a Problem Please (1 Viewer)

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bengaltiger14

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?

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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

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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

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bengaltiger14

So t=VoY/2g. But how do I find that if I do not know the angle?? You cant 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.)

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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.

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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 formular for this.

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

bengaltiger14

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

Delzac

bengaltiger14 said:
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

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