Solving Projectile Motion Problems

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Homework Help Overview

The discussion revolves around a projectile motion problem involving a projectile launched from a cliff. The initial conditions include a height of 205 meters and an initial speed of 155 m/s at an angle of 37 degrees. Participants are exploring various aspects of the projectile's motion, including time of flight, range, and maximum height.

Discussion Character

  • Exploratory, Assumption checking, Mathematical reasoning

Approaches and Questions Raised

  • Participants discuss the equations of motion and their application to determine time of flight and range. There are questions about the correct formulation of the equations, particularly regarding the presence of squares in the equations. Some participants are also questioning their calculations and assumptions about the projectile's height and range.

Discussion Status

The discussion is active, with participants providing corrections to each other's equations and calculations. There is a focus on verifying the time of flight and the range of the projectile, with some participants expressing uncertainty about their results. Guidance has been offered regarding the approach to finding the maximum height, but no consensus has been reached on the specific values.

Contextual Notes

Participants are working under the constraints of a homework assignment, which may limit the information they can use or the methods they can apply. There is also a noted confusion regarding the calculations and the use of trigonometric functions in determining the projectile's motion.

Saladsamurai
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I am doing some review with a friend and I am having trouble with a few problems. I think I am making an incorrect assumption somewhere since all of the problems I am having trouble with are similar.

Problem:
A projectile is shot from the edge of a cliff h=205meters above the ground with an initial speed of v_0=155m/s at an angl of 37 degrees with the horizontal.

Equations: v = v_0 + a t x = x_0 + v_0 t + (1/2) a t^2 v^2 = v_0^2 + 2 a \Delta x in both x and y directions.

From this I have written that:
x_0=0
x_f=?
y_0=205
y_f=0
(v_o)_x=155cos37
(v_0)_y=155sin37

(a) Determine the time taken by the projectile to hit a point P at ground level Should it just be y = (v_0)_y t + (1/2) (-g) t^2?

Which gives me a quadratic?
-205=(155sin37)^2t-4.9t^2=21.03s

(b) Determine the Range of the projectile as neasured from the base of the cliff.

(c) At the instant before the projectile hits point P, find the vertical and horizontal components of its velocity (take up and to the right as positive).

(d) Magnitude and direction of velocity (angle made with the horizontal in degrees below the horizontal):

(e) The MAX height above the cliff top thatthe projectile reached:
 
Last edited:
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your equation should be:

-205=(155sin37)^2t-4.9t^2
 
Oops I didn't notice that square... the equation should be:

-205=(155sin37)t-4.9t^2

no square for the 155sin37.
 
Saladsamurai said:
EDIT: So, in my diagram, I should have ground level=-205m=y_f since up os positive. so t= 1775.82 s. Is this what you are getting? It just seems large.

yf = 0. yi = 205

displacement = yf - yi = 0 - 205 = -205.
 
Is range just the deltax which is 2.85km. Correct?
 
Saladsamurai said:
Is range just the deltax which is 2.85km. Correct?

how do you get that? I'm getting range = vcos(theta)*t = 155cos(37)*21.03 = 2.603km
 
So I really just need help on the maximum eight reached by the [rojectile. I know it will be 205+something...so I need to find out how high its y height before gravity takes over...
so I should use (v_f)_y^2=(v_0)_y^2+2a(\deltay) where v_f=0 and solve for delta y right?!
 
Saladsamurai said:
So I really just need help on the maximum eight reached by the [rojectile. I know it will be 205+something...so I need to find out how high its y height before gravity takes over...
so I should use (v_f)_y^2=(v_0)_y^2+2a(\deltay) where v_f=0 and solve for delta y right?!

yup, that should do it. remember that they only ask for the height above the cliff...
 
learningphysics said:
how do you get that? I'm getting range = vcos(theta)*t = 155cos(37)*21.03 = 2.603km
I don't know..it has to do with the order the calculator does it in...I am putting in like: 155cos(37) ENTER then ANS*23.03=2850m Isn't that correct? It must be...I just tried like this too: (155cos(37))(23.03)=2850m

So for Max height=439.46 meters.

Thanks learnigphysics! review is killing me! It's been about a year since I looked at PMotion...this was a good problem!

Casey
 
Last edited:
  • #10
Shouldn't the time be 21.03s, not 23.03s...

I get height = (155sin37)^2/(2*9.8) = 443.95m

close to your answer... probably just the rounding that's making the difference.
 

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