Hi, im trying to find the initial velocity of this question: Using a video camera, it is observed that when a ball is kicked from A, it just clears the top of a wall at B as it reaches its maximum height. Knowing the distance from A to the wall is 20m and the wall is 4m high, determine the initial velocity at which the ball was kicked. Neglect size of the ball ( Ans Ua = 23.9m/s). So im thinking to find the angle you have to use trig : so 20^2 + 4^2 = square root of 416 = 20.4m (hypotenuse) using Sin, 4/ 20.4 = Invert Sin 0.196 = 11.3 degrees angle the ball is projected. Now i have then angle what else do i need to find in order to find the initial velocity at which the ball was kicked? Thanks for any help
That angle is not the angle of the ball's initial velocity. The ball travels a parabolic trajectory, not a straight line. Analyze the vertical motion first. Figure out the vertical component of the initial velocity.
Ah ok, i thought about the ball not traveling in a straight line assuming the angle was wrong at some point. Ok so if 'm trying to analyze the vertical component of the initial velocity im going to use equation: Sy = Uyt + 0.5(ay)t^2 Or Vy^2 = Uy^2 - 2g(Sy) The 2nd equation i'm thinking i should use since it has just one unknown (Uy). Vy = 0, Sy = 4m 0 = Uy^2 - 2(-9.81)(4) I dont want to go further until i know what i'm doing so far is correct..
Almost perfect. You have an extra minus sign in your last equation. g is just the magnitude of acceleration, so g = 9.81, not -9.81. (The acceleration is -g = -(9.81) = -9.81 m/s^2.)
Ok so Uy^2 = 2(9.81)(4) + 0 Uy^2 = square root 78.48 Uy =8.86 m/s Correct? What shall i do now? Do i find the time? Sy = Uyt + 0.5(9.81)t^ Sy = 4 + (8.86)t - 4.905t^2 Sy = 12.86t / 4.905^2 4 = 12.86 - 4.905 4 = 7.955 t = square root 0.5 = 0.7s t = 0.7 s .. is this correct so far?
Yes. Good. Right approach, but you messed up the equation: (1) 4 is the final position, not the initial. (2) a + bt ≠ (a + b)t Even easier: What's the average vertical speed? Use that to find the time.
No. The vertical speed varies uniformly from its initial value of 8.86 m/s to 0 at the highest point. What do you think the average vertical speed would be?
sice it highest point is Sy = 4m is it 8.86 / 4 = 2.215 m/s average vertical speed? if its not can you provide further help? thanks
No. You cannot divide a speed by a distance and expect to end up with a speed. Try this analogy: You start out making zero money and, uniformly, little by little, end up making $100/day. Over that time period, what was your average salary per day? You may want to review some of the basic formulas of kinematics: Basic Equations of 1-D Kinematics Kinematic Equations and Problem-Solving And here's some good stuff about projectile motion: Projectile Motion
Good! So now that you have the average vertical speed and the vertical distance, figure out the time.
with the analogy: Say you made 0$ to 100$ over 7 days .. average income per day is 100 / 7 = 14.3$ per day
Question: Why does the AVERAGE Initial vertical velocity have to be used instead of just the initial vertical velocity so i can understand why?
No. Your income increases from 0/day to 100/day. Like this: 0/day; 1/day, 2/day, 3/day...... 100/day. Average (assuming uniform increase--just like uniform acceleration) is (0 + 100)/2 = 50/day. Half the time you made less, half the time you made more. (But don't worry about that analogy if it doesn't help.)
No, using that method you would use the initial velocity, not the average velocity. You'll end up with a quadratic to solve. Again, using that method you would use the initial velocity. I wanted to show you an easier way to get the time, using Distance = ave speed X time. No quadratics to solve. (Do it both ways and compare the answers.)
Yes i realize using quadratics is a way to find time, i didn't know there was an easier way to solve for t thou. as for getting the average vertical initial velocity .. i don't know what this answer goes into which equation in order to solve t. Sorry i haven't solved for initial velocity before as this topic (projectile motion) is new to me. Easier methods are always better :) solving for Ux i would have to find time t?
You should solve it both ways to make sure you understand them. I gave you the needed equation in my last post--in words. (Another version is in the first link I gave you.) You're doing OK. Keep at it! Yes. To solve for Ux, you'll first need the time. Which is what you're solving for now.