# Problem about projectile motion

1. Sep 14, 2010

### issacnewton

hi

i have this problem from serway's book. i have posted problem and its solution (from solution manual which i have) in 1.doc my own solution is in r.pdf. i have explained my problem there.

newton

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2. Sep 14, 2010

### alphysicist

Hi issacnewton,

I have not checked all your equations, but I don't believe eq. 15 from your notes shows that the rock never touches the hemisphere. Have you tried plugging in values for x between 0 and R, and finding the position of the rock?

Alternatively, you could write down the equation of a circle, and find where the parabola intersects the circle for x between 0 and R.

3. Sep 14, 2010

### issacnewton

hi

no i have not done the plugging between the endpoints. but at x=0, y is R and at y=0, x is 1.21R, so when the ball lands on the ground it lands beyond the hemisphere, also eq 15 is a parabola. if ball ever touches the hemisphere, how can we get that at y=0, x is 1.21R.
i have known the other approach shown by you, which is also there in the author's solution
posted by me. i am just trying to understand my solution.

thanks

4. Sep 14, 2010

### alphysicist

Let's consider your first case, which gave your Eq. 10. The problem is not that the path is touching the hemisphere at y=0, it is that the entire parabolic path up to that point is inside the rock.

So if you'll calculate some sample points for Eq. 15, I believe you'll find that although it ends up well past the hemisphere, the majority of that trajectory is also inside the rock. For example, set R=1, x=0.5, and find y; then use that to calculate x^2 + y^2. If that result is less than R^2=1, then that part of the path is inside the rock and the path is not viable.

If you think about how the points of intersection between the parabola and hemisphere behave as the initial speed is increased, I think you'll see why the book's answer is the lowest possible speed.

5. Sep 15, 2010

### issacnewton

Eureka....Wow.. Al ... I never considered the possibility that eq 10 could describe a parabola lying entirely inside the hemisphere. i never did numerical calculations..... and the eq 15, its so close to the answer but at x=0.2 R , x^2+y^2 = 0.98 R^2, so its still inside the hemisphere. so even that is not the solution. even if i was wrong, my struggle gave me great insight in the problem, which is necessary in problem solving... now i will look at the solution given by serway....

thanks

6. Sep 15, 2010

### alphysicist

Sure, glad to help!