# Projectile Motion of a Shot Problem

1. May 23, 2010

### hanlon

1. The problem statement, all variables and given/known data

A projectile is shot from the edge of a cliff 115 above ground level with an initial speed of 65.0 at an angle of 35.0 with the horizontal, as shown in the figure .

At the instant just before the projectile hits point P, determine the horizontal and the vertical components of its velocity.

2. Relevant equations

basic kinematic equations [a= constant]

3. The attempt at a solution

My attempt was pretty basic, I thought it was correct, and even asked my brother whos a physics tutor for help and he got the same answer, though the MasteringPhysics site which my class is using says its wrong

the horizontal x direction I got the same speed as it is when shot from the start so it doesn't change

65m/s*cos(35o) = 53.2 m/s in the horizontal direction

for the vertical I used the equation

v = vo + at

where vo = 65m/s*sin(35o) = 37.3 m/s
a = 9.8 m/s2
t = 2.4 s

"t" I found in a different part of the question.

so "v" = 60.36 m/s

though the site says its wrong (masteringphysics)

Thanks.

2. Relevant equations

2. May 23, 2010

### PhysicsTurkey

Use the formula V = (U^2 + 2gh)^1/2
to find the final velocity

3. May 23, 2010

### hanlon

I dont think I've seen that formula before, where you get it? and U is velocity initial right?

also its still equals 60.38, which I already used as a former answer, does that mean my horizontal answer is wrong or the teacher who posted this question online is wrong?

4. May 23, 2010

### PhysicsTurkey

hmm.. It looks like your teacher may have made a mistake then.

and yes "U" is initial Velocity

5. May 23, 2010

### hanlon

there is no answer, its a site called mastering physics, he posts questions and we input an answer if its right we get credit if wrong we lose.

6. May 24, 2010

### rl.bhat

Can you show your calculation of t?
I am using the formula

-115 = 37.3(t) - 0.5*9.8(t)^2.

Solve the quadratic to find t.

7. May 24, 2010

### hanlon

115 = 0 + (3.73m/s)*t + 4.9t2

since velocity and acceleration are moving in the same direction at this point

-37.3 (+/-)sqr(37.32 - 4(4.9)(-115))/ 2*4.9

which ended up 2.355102041 and some other number I didn't record.

8. May 24, 2010

### rl.bhat

115 = 0 + (3.73m/s)*t + 4.9t^2

The displacement and the acceleration are in the same direction.

But the velocity is in the opposite direction to acceleration. So the equation should be

115 = 0 - (3.73m/s)*t + 4.9t2

9. May 24, 2010

### hanlon

ah, I got a email back from my teacher

since its going downwards the velocity component is negative. woops