I feel like an idiot for not being able to figure this one out for myself, but if someone could help me out it would be greatly appreciative.
A home run is hit in such a way that a baseball just clears a wall 25m higher, located 140m from home plate. The ball is hit at an angle of 35 degrees to the horizontal, and air resistance is negligible. Find (a) the initial speed of the ball, (b) the time it takes the ball to reach the wall, and (c) the velocity components and the speed of the ball when it reaches the wall. (assume that the ball is hit at a height of 1.0m above the ground use only the kinematic equations and not any derived formulas.
Not sure if this is right but look it over (if not I’m sorry…i tried)
PART A
Use the formula (Vf)2= (Vi)2 + 2(a)(y)
Vf=final velocity
Vi= initial velocity
a= 9.8 m/s2
y= distance in the vertical (up down) direction
**I think once the ball is clearing the wall at 25 meters high, the velocity will be 0, so assuming that…)
(0)2= (Vi)2 + 2(9.8)(25)
**doing the math and solving for Vi, I got: 22.12 m/s
PART B
Time to reach the wall is just:
time= distance/velocity
To get the x direction velocity use the inital velocity from PART A:
Velocity X = 22.12m/s (cos 35) = 18.14 m/s
Next plug that into the time t=d/v
time= 140 m / 18.14 m/s
time=approx. 8 seconds (7.7)
PART C
I am not sure I understand what they are asking…the velocity in the X direction straight to the wall is the same the whole time so if the X direction velocity is 18.14 m/s intially when hit then it is the same when it hits the wall.
Sorry I am such an idiot but I tried…let me know if you get the right answer or if I was on the right track!
Thanks man, it looks good to me except for the fact that it says the balls starting point is 1 foot, meaning you should use 24 meters instead of 25. I thank you again, you might have earned me a couple extra credit points.
[quote]playmaker08 wrote:
use only the kinematic equations and not any derived formulas.[/quote]
This is the important part, from a pedagogical point of view. The whole point of problems like this is to derive the expressions you need. So make sure you know how to derive the velocity formula, for example.
You won’t be able to solve it using only kinematic equations.
Luckily, you can ignore air friction, so you know the ball travels in a parabola.
The ball starts at 0,0; travels though a vertex, then goes through 140,24.
You need to find the vertex. To do this, you need an equation for the curve. It will be of the form y=ax^2+bx+c. You should be able to find c rather easily.
You have two points, one of them being a root. You also know the line tangent to the curve at (0,0) is 35 degrees. Therefore you can find the slope of the tangent line, a and b, and finally an equation for the parabola.
With this you can find the vertex, which gives you the highest point the ball travels. After this you should be able to find the rest of the information using the kinematic equations.
[quote]playmaker08 wrote:
I feel like an idiot for not being able to figure this one out for myself, but if someone could help me out it would be greatly appreciative.
A home run is hit in such a way that a baseball just clears a wall 25m higher, located 140m from home plate. The ball is hit at an angle of 35 degrees to the horizontal, and air resistance is negligible. Find (a) the initial speed of the ball, (b) the time it takes the ball to reach the wall, and (c) the velocity components and the speed of the ball when it reaches the wall. (assume that the ball is hit at a height of 1.0m above the ground use only the kinematic equations and not any derived formulas.[/quote]
If we assume no friction, the horizontal velocity should be constant until the ball hits something. The vertical velocity should vary according to
-9.8m/(sec^2) or 9.8m/(sec^2) depending on which orientation is used for positive vertical velocity.
If the vertical velocity did not vary, the ball would hypothetically continue upward at an angle of 35 degrees as it passes the wall. This means that absent the downward acceleration, the ball would pass over the wall X meters higher than its starting height, where X/140 = tangent of 35 degrees, if I remember the definition of tangent correctly.
This means the no-gravity increase over its starting height at that point would =
140 * tan(35deg)
so the no-gravity clearance over the wall would be
(140 * tan(35deg)) - 24
The fact that the ball barely clears the wall means that the time it took the ball to travel from its starting point to the wall is just enough time for the ball lose
((140 * tan(35deg)) - 24) meters of height compared to the height the ball would have had with no gravity pulling down on it. I am pretty sure that should be the same amount of time it would take an object starting at rest to fall ((140 * tan(35deg)) - 24) meters.
Let T = the amount of time it would take an object starting at rest to fall ((140 * tan(35deg)) - 24) meters.
Since we are considering distance downward as a positive number and time as positive, for this portion we should consider acceleration caused by gravity to be
positive 9.8m/(sec^2) .
I am pretty sure for an object starting at rest
time = square root of ((2*distance)/acceleration)
– so if I may take the liberty of representing square root as ‘^0.5’, then T referred to in #4 above should be
((2*((140*tan(35deg))-24))/9.8)^0.5
I believe T calculated from #5 above is the time it takes for the ball to travel to the wall. The horizontal component of the ball’s velocity is
140 divided by T meters-per-second
(140/T) m/sec
Picture a right-triangle with a horizontal leg 140 meters, and the hypotenuse angled upward at 35 degrees from that leg. The length of that hypotenuse =
(140/cos(35deg)) .
I think the ball’s initial velocity is
((140/T)/cos(35deg)) m/sec
– or, same value as above –
(140/(T*cos(35deg))) m/sec
where T is the time it takes the ball to reach the wall, from #4, #5, and #6 above.
Assuming #5 above is correct and T actually is
((2*((140*tan(35deg))-24))/9.8)^0.5
then I think the ball’s initial velocity should be
[quote]NealRaymond2 wrote:
This means the no-gravity increase over its starting height at that point would =
140 * tan(35deg)
so the no-gravity clearance over the wall would be
(140 * tan(35deg)) - 24
[/quote]
You’re working way too hard. It looks like Chewie got it right. No need to calculate where the ball would be in the absence of gravity. The important thing to remember is that the vertical and horizontal velocity components are Vcos(35) and Vsin(35); the temptation is to treat them as independent variables Vx and Vy, which obscures the fact that you have two equations and two unknowns.
[quote]Flop Hat wrote:
“Despite a fearsome appearance and temperamental disposition, Wookiees are shown to be very intelligent and technologically proficient.” (wiki)[/quote]
[quote]NealRaymond2 wrote:
playmaker08 wrote:
I feel like an idiot for not being able to figure this one out for myself, but if someone could help me out it would be greatly appreciative.
A home run is hit in such a way that a baseball just clears a wall 25m higher, located 140m from home plate. The ball is hit at an angle of 35 degrees to the horizontal, and air resistance is negligible. Find (a) the initial speed of the ball, (b) the time it takes the ball to reach the wall, and (c) the velocity components and the speed of the ball when it reaches the wall. (assume that the ball is hit at a height of 1.0m above the ground use only the kinematic equations and not any derived formulas.
If we assume no friction, the horizontal velocity should be constant until the ball hits something. The vertical velocity should vary according to
-9.8m/(sec^2) or 9.8m/(sec^2) depending on which orientation is used for positive vertical velocity.
If the vertical velocity did not vary, the ball would hypothetically continue upward at an angle of 35 degrees as it passes the wall. This means that absent the downward acceleration, the ball would pass over the wall X meters higher than its starting height, where X/140 = tangent of 35 degrees, if I remember the definition of tangent correctly.
This means the no-gravity increase over its starting height at that point would =
140 * tan(35deg)
so the no-gravity clearance over the wall would be
(140 * tan(35deg)) - 24
The fact that the ball barely clears the wall means that the time it took the ball to travel from its starting point to the wall is just enough time for the ball lose
((140 * tan(35deg)) - 24) meters of height compared to the height the ball would have had with no gravity pulling down on it. I am pretty sure that should be the same amount of time it would take an object starting at rest to fall ((140 * tan(35deg)) - 24) meters.
Let T = the amount of time it would take an object starting at rest to fall ((140 * tan(35deg)) - 24) meters.
Since we are considering distance downward as a positive number and time as positive, for this portion we should consider acceleration caused by gravity to be
positive 9.8m/(sec^2) .
I am pretty sure for an object starting at rest
time = square root of ((2*distance)/acceleration)
– so if I may take the liberty of representing square root as ‘^0.5’, then T referred to in #4 above should be
((2*((140*tan(35deg))-24))/9.8)^0.5
I believe T calculated from #5 above is the time it takes for the ball to travel to the wall. The horizontal component of the ball’s velocity is
140 divided by T meters-per-second
(140/T) m/sec
Picture a right-triangle with a horizontal leg 140 meters, and the hypotenuse angled upward at 35 degrees from that leg. The length of that hypotenuse =
(140/cos(35deg)) .
I think the ball’s initial velocity is
((140/T)/cos(35deg)) m/sec
– or, same value as above –
(140/(T*cos(35deg))) m/sec
where T is the time it takes the ball to reach the wall, from #4, #5, and #6 above.
Assuming #5 above is correct and T actually is
((2*((140*tan(35deg))-24))/9.8)^0.5
then I think the ball’s initial velocity should be
Ok, so who’s good at biochem? Cause I need someone to explain the inhititors of PDH, TCA, Glycolysis, gluconeogenisis, ETC, and beta ox by tomorrow morn. Actually I think I’ll get a coffee and head for the lib.