Angled Projectile Motion Question

[xxlvh]

1. An object was thrown up at an angle of 22^o from the horizontal. If this object travels 38 m horizontally, what was the speed at which the object was thrown?

angle = 22^o
d_horizontal = 38m
g = -9.81m/s²

I have absolutely no clue on how to even go about starting to solve this question...would anyone be able to give me a small hint perhaps on which formula to work with?

 

[rock.freak667]

Put the intial speed as V inclined at the angle of 22. Split that into vertical and horizontal components. (Note that the horizontal velocity is constant).
Then use your kinematic equations.

 

[xxlvh]

If I have the angle and split it into the x and y components, wouldn't I still need at least one velocity given to solve for any of the other sides?

 

[rock.freak667]

Find the total time for the motion ($s=ut+\frac{1}{2}at^2$) in terms of V.

Then the total horizontal distance travelled=Horizontal Speed * Total time for the motion

 

[xxlvh]

Hmm. The formulas I used were just slightly different, I've never seen that one before...what would the variables of "s" and "u" stand for?

 

[rock.freak667]

Hmm. The formulas I used were just slightly different, I've never seen that one before...what would the variables of "s" and "u" stand for?

s=displacement
u=initial velocity

 

[xxlvh]

Okay, thank you.
And Vi would be...22?
Sorry for all the questions. I wasn't having any problems in the whole unit up until this last question!

 

[rock.freak667]

Okay, thank you.
And Vi would be...22?
Sorry for all the questions. I wasn't having any problems in the whole unit up until this last question!

Vi=initial veritcal velocity=Vsin22.

 

[xxlvh]

Alright...I've gotten it to:

t = -2(Vsin22^o) / -9.81m/s²

 

[rock.freak667]

Alright...I've gotten it to:

t = -2(Vsin22^o) / -9.81m/s²

Good.

Now consider horizontal motion.

Horizontal Distance=Horizontal speed * Time


and you have t=(2Vsin22)/9.81 and the horizontal distance is 38

 

[xxlvh]

Okay...So 38 m = V x (2vsin22/-9.81)

Are you able to combine the two V's, if one is horizontal and one is vertical?

When I tried that in the next step I have -372.78m=2v²sin 22^o...when working through it though, it ends up with the root of a negative number so I'm assuming I made a mistake somwhere. =/

 

[rock.freak667]

Okay...So 38 m = V x (2vsin22/-9.81)

Are you able to combine the two V's, if one is horizontal and one is vertical?

When I tried that in the next step I have -372.78m=2v²sin 22^o...when working through it though, it ends up with the root of a negative number so I'm assuming I made a mistake somwhere. =/

$$\frac{-2(Vsin22)}{ -9.81}$$

minus signs cancel out.

and Horizontal velocity=Vcos22.

 

[xxlvh]

Solving for V, I got 23 m/s. According to the book that is the correct answer!

This will add 3% onto my overall mark :) Thank you, I really owe you one!

 

[rock.freak667]

Once you understand how to do the problems is all that's needed.