Solving Projectile Motion Problems

[Saladsamurai]

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 $$v_0=155$$m/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:

 

[learningphysics]

your equation should be:

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

 

[learningphysics]

Oops I didn't notice that square... the equation should be:

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

no square for the 155sin37.

 

[learningphysics]

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.

 

[Saladsamurai]

Is range just the deltax which is 2.85km. Correct?

 

[learningphysics]

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

 

[Saladsamurai]

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?!

 

[learningphysics]

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...

 

[Saladsamurai]

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

 

[learningphysics]

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.