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### Course: Algebra 1>Unit 14

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.

## Want to join the conversation?

• What if you're trying to find the maximum height and not how long it's in the air?
• You'd want to find the vertex of the parabola. See the graph I made in the question above for an approximate answer.
• where did the 16t^2 come from?
• Where did you get -32x and why that number
• What would you do if you were using the quadratic formula and came up with a negative square root?
• 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.
• How would this problem be solved using the "completing the square" method instead of the quadratic formula?
• 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?
• 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!)

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.
• 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.
• 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.
Just look:
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
• On Earth, acceleration due to gravity is 9.8 m/s². This corresponds to 32.2 ft/s².
• 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?
• Yes, you can do that. The equation has a variable for height (h), so you could plus in any value that you want.
• Can the problem be solved by using completing the square? Why?
• 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.