Sal solves a word problem about a ball being shot in the air. The equation for the height of the ball as a function of time is quadratic. Created by Sal Khan and Monterey Institute for Technology and Education.
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- What if you're trying to find the maximum height and not how long it's in the air?(2 votes)
- You'd want to find the vertex of the parabola. See the graph I made in the question above for an approximate answer.(41 votes)
- where did the 16t^2 come from?(19 votes)
- What would you do if you were using the quadratic formula and came up with a negative square root?(12 votes)
- Then you would say that there are no real solutions to the quadratic equation. But when you get into higher levels of math, you will come across imaginary numbers, and there will be a solution. But for now, just say that there aren't any real solutions.(19 votes)
- How would this problem be solved using the "completing the square" method instead of the quadratic formula?(16 votes)
- Actually, the quadratic formula is derived from the completing the square method and so they are essentially, the same. They always work. quadratic formula derivation video
- We could take h to be 0 or -50. Dependent on perspective. Since there are 2 different perspectives, how do we determine whether it's 0 or -50?(5 votes)
- Often the simplest perspective is the one to use, and there may be clues in the words that are used. A good technique is to try to sketch the circumstances in the problem and then think carefully about what's happening.
The problem says you are 50 feet ABOVE the ground. So in a drawing, that point would normally be +50.
The problem also asks how long will it take for the ball to HIT THE GROUND. So, after "t" seconds, it has come back down to earth with no "height" remaining. That's why Sal used zero in the equation.
I hope this is helpful! (I've never found word problems easy!)(10 votes)
- Does be gravitation acceleration in this exercise 32m/s^2 ?
I know its a exercise about quadratic equation not kinematic, but i it can confuse.
s = Vavg. * Δt = (Vi + Vf)/2 * Δt = (Vi + Vi + a*Δt)/2 * Δ = Vi*Δt + (a*Δt^2)/2
then we will subtitute the variables with values
Vi = 20m/s, a = -32m/s^2 (beacuse its accelerating to the ground), s= -50m (negative because we need 50 meters to reach the ground)
-50m = 20m/s * Δt + (-32m/s^2 * Δt^2)/2
-50m = 20m/s * Δt - 16m/s^2 * Δt^2
0 = 20m/s * Δt - 16m/s^2 * Δt^2 + 50m // if we leave the Δ symbol, and m/s we will get...
0 = -16t^2 + 20t + 50(5 votes)
- Can someone please help me understand this problem?:
A ball is thrown directly upward from a height of 30 feet with an initial velocity of 64 feet per second. The equation h=-16t^2+64t+30 gives the height h after t seconds.
Solve for both t and h.(5 votes)
- Maximum height h -> find the axis of symmetry -> x= -b/2a= -64/2(-16)= 2, plug it in -> -16(2)^2+64(2)+30= 94, so maximum height is 94 ft.
At what time will the ball reach the ground? Set h equal to 0 -> 0=-16t^2+64t+30, solve for t, +- means plus or minus -> (-64+-sqrt(64^2-4*-16*30))/2*-16, t=-0.424s and t=4.424s. Since time cannot be negative, it will take the ball 4.424 seconds to hit the ground.(4 votes)
- What if I would want to know at what time would the ball pass certain point?
For example, if I would want to know at what time would the ball pass the same height--going down-- as the initial launch?
Would an equation like 50 = - 16t^2 - 20t + 50 give me an answer?(4 votes)
- Yes, you can do that. The equation has a variable for height (h), so you could plus in any value that you want.(4 votes)
- Can the problem be solved by using completing the square? Why?(4 votes)
- Yes you can. Completing the square works on any quadratic equation. It can get take a lot of time to do it but it always works.(2 votes)
- I am working on Solving Applications Involving Distance, Rate, and Time, (in a word problem) I am having a very hard time for some reason grasping how to turn time into a fraction. ex: 5 hours and 20 mins. I want to write it as 320/60 but I know that is not right.(2 votes)
- That's correct if the unit is hours, but you need to figure out what unit you want.
5 h + 20 min = 320 min = 320 / 60 h = 16 / 3 h(4 votes)
A ball is shot into the air from the edge of a building, 50 feet above the ground. Its initial velocity is 20 feet per second. The equation h-- and I'm guessing h is for height-- is equal to negative 16t squared plus 20t plus 50 can be used to model the height of the ball after t seconds. And I think in this problem they just want us to accept this formula, although we do derive formulas like this and show why it works for this type of problem in the Khan Academy physics playlist. But for here, we'll just go with the flow on this example. So they give us the equation that can be used to model the height of the ball after t seconds, and then say about how long does it take for the ball to hit the ground. So if this is the height, the ground is when the height is equal to 0. So hitting the ground means-- this literally means that h is equal to 0. So we need to figure out at which times does h equal 0. So we're really solving the equation 0 is equal to negative 16t squared plus 20t plus 50. And if you want to simplify this a little bit-- let's see, everything here is divisible at least by 2. And let's divide everything by negative 2, just so that we can get rid of this negative leading coefficient. So you divide the left hand side by negative 2, you still get a 0. Negative 16 divided by negative 2 is 8. So 8t squared. 20 divided by negative 2 is negative 10. Minus 10t. 50 divided by negative 2 is minus 25. And so we have 8t squared minus 10t minus 25 is equal to 0. Or if you're comfortable with this on the left hand side, we can put on the left hand side. We could just say this is equal to 0. And now we solve. And we could complete this square here, or we can just apply the quadratic formula, which is derived from completing the square. And we have this in standard form. We know that this is our a. This right over here is our b. And this over here is our c. And the quadratic formula tells us that the roots-- and in this case, it's in terms of the variable t-- are going to be equal to negative b plus or minus the square root of b squared minus 4ac, all of that over 2a. So if we apply it, we get t is equal to negative b. b is negative 10. So negative negative 10 is going to be positive 10. Plus or minus the square root of negative 10 squared. Well, that's just positive 100, minus 4 times a, which is 8, times c, which is negative 25. And all of that over 2a. a Is 8. So 2 times 8 is 16. And this over here, we have a-- let's see if we can simplify this a little bit. The negative sign, negative times a negative, these are going to be positive. 4 times 25 is 100, times 8 is 800. So all that simplifies to 800. And we have 100 plus 800 under the radical sign. So this is equal to 10 plus or minus the square root of 900, all of that over 16. And this is equal to 10 plus or minus-- square root of 900 is 30-- over 16. And so we get time is equal to 10 plus 30 over 16, is 40 over 16, which is the same thing if we divide the numerator and denominator by 4 to simplify it as 10 over-- or actually even better, divide it by 8-- that's 5 over 2. So that's one solution, if we add the 30. If we subtract the 30, we'd get 10 minus 30. Or t is equal to 10 minus 30, which is negative 20 over 16. Divide the numerator and the denominator by 4, you get negative 5 over 4. Now, we have to remember, we're trying to find a time. And so a time, at least in this problem that we're dealing with, we should only think about positive times. We want to figure out how much time has taken-- how long does it take for the ball to hit the ground? We don't want to go back in time. So we don't want our negative answer right here. So we only want to think about our positive answer. And so this tells us that the only root that should work is 5/2. And we assume that this is in seconds. So this is in 5/2 seconds. I wouldn't worry too much about the physics here. I think they really just want us to apply the quadratic formula to this modeling situation. The physics, we go into a lot more depth and give you the conceptual understanding on our physics playlist. But let's verify that we definitely are at a height of 0 at 5/2 seconds, or t is equal to 5/2. This expression right over here does give us h is equal to 0. So we have-- let's try it out. We have negative 16 times 5/2 squared plus 20 times 5/2 plus 50. This needs to be equal to 0. So this is negative 16 times 25/4 plus-- let's see, if we divide 20 by 2, we get 10. If we divide 2 by 2, we get 1. So 10 times 5 is going to be 50. Plus 50. This needs to be equal to 0. Negative 16 divided by 4 is negative 4. 4 divided by 4 is 1. So you have negative 4 times 25, which is 100. Plus 50-- oh, sorry. Negative 4 times 25 is negative 100. Plus 50, plus 50 again is equal to 0. And so we have negative 100 plus another 100. Well, that's definitely going to be equal to 0. We get 0 equals 0. And it all checks out. We hit the ground after 5/2 seconds. Or another way to think about it is 2.5 seconds. t is equal to 5/2 seconds, or 2.5 seconds.