What is the initial velocity of the projectile given the displacement and angle?

AI Thread Summary
To determine the initial velocity of a projectile given a vertical displacement of 100 cm, a horizontal displacement of 200 m, and an angle of 15 degrees, kinematic equations are essential. The discussion highlights confusion over using the correct angle in calculations, as some participants mistakenly referenced 60 degrees instead of 15 degrees. The quadratic equation formed from the motion equations yields two potential initial velocity values, but there is uncertainty about their validity. It is clarified that the projectile's point of projection affects the initial velocity, indicating that multiple velocities could satisfy the conditions. Ultimately, the participants reach a better understanding of the problem and the necessary calculations.
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Homework Statement


i have a problem from an in class lab where i have to find the initial velocity of the ball. given is the horizontal and vertical displacement of the projectile.
Dy(height)=100cm
Dx(range)=200m
0(theta)= 15degrees.
Find the initial velocity.

Homework Equations


not sure what equation to start with

The Attempt at a Solution


x-comp:
Vx1=Vo cos 60

y-comp:
Vy1=Vo sin 60

t=200/Vo cos 60

-100=Vo sin 60(200/Vo cos 60) + 1/2(-9.8)(200/Vo cos 60)^2
Vo= 2 numbers that make no sense
 
Last edited:
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Draw the initial velocity v0 at the angle 15 and then split that into vertical and horizontal components.

Then use your equations of motion.
 
hey
i haven't learned equations of motion only kinematic equations
 
Last edited:
helpmee said:
hey
i haven't learned equations of motion only kinematic equations

That's what he meant, they are the same thing. Use the kinematic equations.
 
i still don't understand
can u be specific on which ones to use?
 
helpmee said:
i still don't understand
can u be specific on which ones to use?

the working in your original post looks fine, what values for v0 did you get? Also I don't understand how you have sin60 and cos60 when your angle is 15. Also the '-100' should just be 100
 
sorry that should be 15 not 60 and in class we learned that dy was also negative in this type of question
 
helpmee said:
sorry that should be 15 not 60 and in class we learned that dy was also negative in this type of question

it was a lab, so I assume, you all weren't at the ground just shooting it at 15 degrees but did it ona a table at 100cm. In that case, you are right, it is -100.

So forming a quadratic in v0 what values of v0 did you get?
 
i got 2000m/s and 0.00125m/s. i think i might have done the calculations wrong though
 
  • #10
helpmee said:
i got 2000m/s and 0.00125m/s. i think i might have done the calculations wrong though

yes check that back, because your equation shows you will get something like av02-b=0
 
  • #11
can u help me on how to approach the math on this question?
 
  • #12
ok so this what you have right?


-100=(V_o sin15)*(\frac{200}{V_o cos15}) + \frac{1}{2}(-9.8)(\frac{200}{V_o cos 15})^2


So in the first term there are we seeing anything canceling out? Can we simplify it if anything did cancel out?


in the second term, when you square out everything, can you find the number K such that the second term is K/Vo2 ?
 
  • #13
In the problem D(y) is 100 cm. Is it correct?
Mainly point of projection is not given. If it is from the ground, at a given angle of projection and range, D(y) will be attained at different time depending on the initial velocity. So there cannot be a unique value of initial velocity.
 
  • #14
Thanks for the help. i got it now
 
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