Hi, I am new here, and am not expecting any direct answers (I read the sticky) but I am pretty confused about this one question: A ball is shot vertically upwards at 10m/s from a height of 5m above the ground, If friction can be ignored, what is the speed of the ball immediately before it strikes the ground? I need to know if the formula I am using is correct: Vf= Vi+g.T would g (9.8m/s2) be -9.8m/s2 for this question because it uses deceleration? thanks alot!
Welcome to PF! It is certainly A correct formula, and if you choose Vi to be positive (that is, upwards direction positive), you should definitely use the negative -9.8. However, as yet, T is also an unknown, not only Vf! So, (it seems..) you need yet another equation in order to close your system. Do you have any idea as to what that might be?
Your formula is correct; the acceleration due to gravity is constant, so the total change in acceleration from launch to landing is gt. However, you're going about this problem the hard way, since you'd need to find t first to use your formula, and finding t in this case could be a bit complicated. How about using an energy argument instead? The ball begins with a specific amount of kinetic energy, which it trades for altitude as it rises to the top of its trajectory. It then begins to fall, and trades that altitude back into kinetic energy. Since there is no friction, no energy is lost. The ball is going the same speed when it was launched upwards from 5 m altitude as when falling downward past 5 m altitude. In other words, it's going 10 m/s downwards as it passes 5 m altitude. Now, the problem is significantly simplified! All you need to do is determine how much speed the ball would gain in falling the additional 5 meters to the ground, then add it to 10 m/s it had at 5 m altitude. This doesn't require time at all; all you need to do is figure out how much gravitational potential energy the ball lost in moving downwards 5 meters ([itex]= mg\Delta h[/itex]). All of that gravitational potential energy is converted into kinetic energy (so [itex]mg\Delta h = (1/2) m(\Delta v)^2[/itex]). Can you solve that equation for [itex]\Delta v[/itex]? Hint: This is an extremely common sort of thing to be asked in a high-school physics class -- how much speed does a ball gain in falling x meters. You might want to remember the very simple formula you'll find. - Warren
I fully concur with chroot here; an energy argument is definitely the simplest (I gave an oblique hint to that). However, if you are at the beginning of learning physics, it can be quite helpful to verify that energy arguments and a plodding solution of Newton's second law directly gives the same answers (which must "obviously" be true)
Finding the time (if you decide to go that route) involves solving the position equation: [tex]y(t) = y_0 + v_0 t + \frac{1}{2} g t^2[/tex] where [itex]y_0 = 5[/itex] and [itex]v_0 = 10[/itex]. - Warren
I think I am going to use both methods so I am sure that I am correct, it will take a while, but at least I will know how to answer this type of question in the future! I will be sure to remember what a energy argument is! thanks again!
rmackay: Feel free to post your intermediate results for us, so we can make sure you stay on the right track. - Warren