ryan g Posted April 16, 2009 Share Posted April 16, 2009 Hi first post on here - hope someone can help If I were to throw a ball in the air and the ball left my hand at: a speed of 7 m s -1 and at a height of 1.6 m above the ground how high would the ball go? what equation should be used to work this out? assuming the acceleration due to gravity is 9.8 m s -2 any help would be grand many thanks R Link to comment Share on other sites More sharing options...
Cap'n Refsmmat Posted April 16, 2009 Share Posted April 16, 2009 It'll come in two parts. First you need to calculate how long it will take for the ball to decelerate to 0 m/s, because 0m/s will occur only when the ball is at the top and it stops going up and starts going down. After that, you take that time and use it in your good ol' [math]x = x_0 + v_0t + \frac{1}{2} at^2[/math] equation (or whatever version of it you've been taught). That can give you the maximum height. Link to comment Share on other sites More sharing options...
swansont Posted April 16, 2009 Share Posted April 16, 2009 Alternately, you can use the equation [math]2\vec{a}\cdot\vec{s}=v_f^2-v_i^2[/math] (the dot product means that if the vectors are in opposite directions, there will be a - sign) Link to comment Share on other sites More sharing options...
Cap'n Refsmmat Posted April 16, 2009 Share Posted April 16, 2009 Although I learned that one as [math]v_f^2 = v_i^2 + 2a(x_f - x_i)[/math], and we ignored the vectors. Weird how different people present the equations. And yeah, I forgot that'd work. Whoops. Link to comment Share on other sites More sharing options...
ryan g Posted April 16, 2009 Author Share Posted April 16, 2009 OK - what if I wanted to find the speed that the ball is travelling when it hits the floor? The question I need to answer simply requires the speed on impact with the ground. Im a bit worried about working out how long the ball will take to reach its maximum hieght because Im not sure ive been taught how to do that yet and not sure if its necessary for my answer Is it possible to work out the speed the ball is travelling when it hits the ground WITHOUT knowing the maximum height the ball reaches in the air? I have worked out an answer but the speed is lower than that of which the ball was travelling when it was thrown. Im guessing thats incorrect due to acceleration due to gravity? any help is greatly apprieciated ! Link to comment Share on other sites More sharing options...
Cap'n Refsmmat Posted April 16, 2009 Share Posted April 16, 2009 Have you learned the equations for potential and kinetic energy yet or are you just doing kinematics? If you use the above equations to find out how high the ball will go, you can use the same ones to find out how fast it will be going when it hits the ground when falling from that height. Link to comment Share on other sites More sharing options...
ryan g Posted April 17, 2009 Author Share Posted April 17, 2009 Ive learned the equations for kinetic and potential energy - no idea about the other. Im still a bit confused - how do I apply them to work out the hieght? sorry if this appears dumb - im returning to education / learning at home for the first time in ten years and its a bit of a nightmare trying to get my head around everything. thank you thank you Merged post follows: Consecutive posts mergedwould I be right in thinking as the ball is travelling at 7 m s when thrown and the gravitational energy is 9 m s the ball will stop travelling at 2 metres above the release point. or is that just wrong? Link to comment Share on other sites More sharing options...
Sisyphus Posted April 17, 2009 Share Posted April 17, 2009 Yes, that's wrong. What you want is to figure out equations for height vs. time, velocity vs. time, and acceleration vs. time. Each of those is a derivative (rate of change) of the one before it. You already have the last one: it's a constant. a = -9.8. What, then, is velocity? You have the initial velocity, 7, and you have it's rate of change with respect to time, -9.8 per second. So what would the equation be, if v is velocity and t is time? Link to comment Share on other sites More sharing options...
ryan g Posted April 17, 2009 Author Share Posted April 17, 2009 I dont think this is right but v = change in speed / time taken 7 = - 9.8 / t x t t7 = -9.8 /7 t = 1.4 i dont think Ive got the right equation. is magnitude of acceleration the same as velocity? really sorry - the answer is probably staring me in the face! Link to comment Share on other sites More sharing options...
Sisyphus Posted April 17, 2009 Share Posted April 17, 2009 Acceleration is the rate of change of velocity, just like velocity is the rate of change of position. So, if the acceleration is -9.8 meters per second per second, then your velocity is going to decrease by 9.8 meters per second for every second, meaning at a given time t seconds after you start, your velocity is going to be 9.8*t less than your initial velocity. And you know your initial velocity, which is 7 meters per second. So the equation for velocity is going to be: v = 7 - 9.8t So you can see, 1 second after you throw the ball in the air, it has the velocity has changed by 9.8, down from 7, meaning it will be -2.8 meters per second. And so on, for any other time. When will the velocity equal zero? Link to comment Share on other sites More sharing options...
ryan g Posted April 17, 2009 Author Share Posted April 17, 2009 7 / 2.8 = 2.5 2.5 secs so after leaving the hand the ball is in the air for 2.5 secs before reaching its maximum height. that seems reasonable - am i getting warmer? Link to comment Share on other sites More sharing options...
swansont Posted April 17, 2009 Share Posted April 17, 2009 The equation I posted is a kinematics equation, but if you multiply by the m/2, you'll see that it's kinetic and potential energy, and can be used to solve both questions. It also works with the version Cap'n Refsmmat posted, of course. Link to comment Share on other sites More sharing options...
ryan g Posted April 17, 2009 Author Share Posted April 17, 2009 sorry - i don't know eithier of those equations and have no idea how to adapt them to get the height. is 2.5 secs correct? so distance travelled = speed x time taken = 7 m s x 2.5 s = 17.5 m then do I do this : 17.5 - 9.8 = 7.7 m 7.7 m + 1.6m = 9.3 m ????????? Link to comment Share on other sites More sharing options...
Cap'n Refsmmat Posted April 17, 2009 Share Posted April 17, 2009 Not quite. It's not 2.5 seconds, for one thing -- you meant 7 / 9.8, not 7 / 2.8. Once you get the correct time, you can't just multiply by 7 m/s -- it's going to be slowing down as it flies through the air because of gravity. If you're familiar with it, you need to use this equation: [math]x = x_0 + v_0t + \frac{1}{2}at^2[/math] where a = the acceleration due to gravity (-9.8), x is the final height, x0 is the initial height, v0 is the initial velocity, and t is time. Link to comment Share on other sites More sharing options...
ryan g Posted April 18, 2009 Author Share Posted April 18, 2009 aha - im getting it now maybe trying to cram the whole book on energy and light in 2 days isn't such a good idea! I have that equation written down somewhere - I'll have a go in the morning and let you know how I get on many thanks for your time and patience!! Link to comment Share on other sites More sharing options...
ryan g Posted April 20, 2009 Author Share Posted April 20, 2009 ok - I plotted a rough graph with velocity against time - deducting 2.8 metres for each second. the time at 0 velocity is roughly 3.5 secs. Is that correct? Merged post follows: Consecutive posts mergedthis is so frustrating - I have all the rest of the answer planned - to get the final speed on impact with the ground - just finding the height and I know its so simple!! argh. I am now using the actual values stated in the question to work it out and the time the ball takes to reach speed 0 m s = 3.9 s (using initial speed of 7.8 m s rather than 7 m s and acceleration due to gravity -9.8 m s ball leaves hand at 1.4m above thr ground.) slotting this in the equation you gave me gives me x = -42.68m !!!!!! surely impossible Link to comment Share on other sites More sharing options...
swansont Posted April 20, 2009 Share Posted April 20, 2009 You may find it easier to use the energy equations, or he related kinematic equations. KE = 1/2 mv^2 and PE = mgh. Once the ball is released, PE + KE = constant, so you can choose the release point to be h=0 (or any other convenient point, as long as you apply it consistently) Divide by m (since it isn't known) and you get v^2/2 + gh = constant Link to comment Share on other sites More sharing options...
ryan g Posted April 20, 2009 Author Share Posted April 20, 2009 Ok I think ive got it! many many thanks for all your help! Merged post follows: Consecutive posts mergedactually I dont got it! the constant i worked out to be 2.38 how do i then apply that number to work out the height? or is that the height? Merged post follows: Consecutive posts mergedhang on thats totally wrong - i used the wrong values. im getting really confused which equations to use - Im not getting taught kinematics. right start from the top. im covered in scraps of paper with all sorts of equations on an am totally lost. Link to comment Share on other sites More sharing options...
swansont Posted April 20, 2009 Share Posted April 20, 2009 v^2/2 + gh = constant v^2/2 = 24.5 m^s/s^2, so at h=0, this is the value the sum must always have. At the maximum height, v=0, so gh= 24.5 m^s/s^2 When it hits the ground, you again solve v^2/2 + gh= 24.5 remember that h= -1.6 m (you've defined h=0 as the release point) Link to comment Share on other sites More sharing options...
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