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Video: Solving Quadratic Equations Involving Projectiles

Bethani Gasparine

The height of a missile above its launch point can be found using 𝑠 equals 𝑢𝑡 minus four point nine 𝑡 squared, where 𝑠 meters is the height above the launch point, 𝑢 meters per second is the vertical launch speed, and 𝑡 seconds is the time after launch. A missile is launched vertically upwards with a speed of 49 meters per second. At what times will it be 44.1 meters above its launch point?

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Video Transcript

The height of a missile above its launch point can be found using 𝑠 equals 𝑒𝑑 minus four point nine 𝑑 squared, where 𝑠 meters is the height above the launch point, 𝑒 meters per second is the vertical launch speed, and 𝑑 seconds is the time after launch. A missile is launched vertically upwards with a speed of 49 meters per second. At what times will it be 44.1 meters above its launch point?

So using this equation, 𝑠 is the height above the launch point which they said would be 44.1 meters. And the speed 𝑒 is 49 meters per second. And now we will be solving for 𝑑.

Let’s go ahead and put this in standard form. So we will add 4.9 t squared to the left and then subtract 49𝑑 also to the left. And now we can take out a greatest common factor. And when we take 4.9, out we have 𝑑 squared minus 10𝑑 plus nine.

So we need numbers that multiply to be nine and add be negative 10. So which of these pairs would add to be negative 10? That would be negative one and negative nine. So our factors would be 𝑑 minus one and 𝑑 minus nine and then a gcf of 4.9. So if we set our factors equal to zero, we get 𝑑 equals one and 𝑑 equals nine.

So the missile is at a height of 44.1 meters at 𝑑 equals one and 𝑑 equals nine seconds.