# Homework Help: Projectile Motion 1

1. Oct 10, 2007

### 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:

Last edited: Oct 10, 2007
2. Oct 10, 2007

### learningphysics

your equation should be:

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

3. Oct 10, 2007

### learningphysics

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

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

no square for the 155sin37.

4. Oct 10, 2007

### learningphysics

yf = 0. yi = 205

displacement = yf - yi = 0 - 205 = -205.

5. Oct 10, 2007

### Saladsamurai

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

6. Oct 10, 2007

### learningphysics

how do you get that? I'm getting range = vcos(theta)*t = 155cos(37)*21.03 = 2.603km

7. Oct 10, 2007

### 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?!

8. Oct 10, 2007

### learningphysics

yup, that should do it. remember that they only ask for the height above the cliff...

9. Oct 10, 2007

### Saladsamurai

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: Oct 10, 2007
10. Oct 10, 2007

### 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.

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