Question Video: Using the Quadratic Formula to Solve Word Problems | Nagwa Question Video: Using the Quadratic Formula to Solve Word Problems | Nagwa

Question Video: Using the Quadratic Formula to Solve Word Problems

A shell was fired from a mortar along a trajectory described by the equation 𝑦 = 0.19 + 0.31𝑥 − 0.5𝑥², where 𝑦 is the height of the shell above the ground in kilometres when it has travelled a horizontal distance of 𝑥 kilometres. Find the horizontal distance covered by the shell before it hit the ground.

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Video Transcript

A shell was fired from a mortar along a trajectory described by the equation 𝑦 equals 0.19 plus 0.31𝑥 minus 0.5𝑥 squared, where 𝑦 is the height of the shell above the ground in kilometres when it has travelled a horizontal distance of 𝑥 kilometres. Find the horizontal distance covered by the shell before it hit the ground.

So if we would fire a shell, it would go up in the air and then eventually hit the ground. And 𝑦 represents the height of that shell, so when it hits the ground, 𝑦 would be zero and 𝑥 represents how far it would travel horizontally. So if we take our equation and we put it in descending order, we would put the 𝑥 squared term first, then the 𝑥, and then the constant. So from here, we will know the distance travelled before it hit the ground and it will hit the ground when 𝑦 is zero. So we said that it’s equal to zero.

So we can solve this quadratic equation a few ways, but the best way since there’s lots of decimals would be the quadratic formula. The quadratic formula is negative 𝑏 plus or minus the square root of 𝑏 squared minus four 𝑎𝑐 all over two 𝑎. 𝑎 is the coefficient in front of 𝑥 squared, 𝑏 is the coefficient in front of 𝑥, and 𝑐 is your constant. So taking our values, we can solve for 𝑥, the distance that it travelled horizontally, by plugging these into that quadratic formula. Now that they’ve been plugged in, let’s evaluate.

So after squaring our decimal and multiplying four 𝑎𝑐, let’s simplify this more. And now let’s take the square root and we get 0.31 plus or minus 0.69. But before we get too far, let’s take a few steps back and look at our signs. The negative 0.31 cancelled with the negative with the one, so that’s why 0.31 is positive. And then since we have a plus and a minus, it wouldn’t matter if we would bring the negative up with it. It would still have a plus and a minus.

So we need to take 0.31 and add 0.69 and also subtract 0.69. So we get 𝑥 equals one or 𝑥 equals negative 0.38. However, this is a horizontal distance, so it’s not going to travel at negative amount. So that means after the shell was fired, it travelled a horizontal distance of one kilometre before it hit the ground.

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