- #1
issacnewton
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IssacNewton said: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.
please read it and try to answer
Newton
alphysicist said: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?
IssacNewton said: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
alphysicist said: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.
IssacNewton said: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
Projectile motion refers to the motion of an object that is projected into the air and then moves under the influence of gravity alone. This type of motion is commonly seen in sports like baseball, basketball, and golf.
The key factors that affect projectile motion are the initial velocity, angle of projection, and the force of gravity. These factors determine the path and trajectory of the object as it moves through the air.
The range of a projectile is calculated using the formula R = (V^2 * sin2θ)/g, where R is the range, V is the initial velocity, θ is the angle of projection, and g is the acceleration due to gravity.
The horizontal component of projectile motion refers to the motion in the horizontal direction, while the vertical component refers to the motion in the vertical direction. The horizontal component of motion is affected by the initial velocity, while the vertical component is affected by the force of gravity.
Air resistance can affect projectile motion by slowing down the object's velocity and altering its trajectory. This can result in a shorter range and a different path compared to a projectile in a vacuum. Air resistance is typically more significant for objects with larger surface areas, such as feathers or parachutes.