# Help with a Problem Please (1 Viewer)

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#### 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

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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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