# Question Video: Determining the Speed of a Body Based on the Expression of Its Height in Terms of Time Mathematics

A stone is projected vertically upward. At time 𝑡 seconds, its height from the ground is given by ℎ = (46.6𝑡 − 4.9𝑡²) m, 𝑡 ≥ 0. Determine the speed of the stone when it is 22.5 m high.

04:29

### Video Transcript

A stone is projected vertically upward. At time 𝑡 seconds, its height from the ground is given by ℎ equals 46.6𝑡 minus 4.9𝑡 squared meters, for 𝑡 is greater than or equal to zero. Determine the speed of the stone when it is 22.5 meters high.

We’re given an equation for ℎ in terms of 𝑡. ℎ is the height of the stone from the ground. In other words, it’s its displacement. Now, the question wants us to find the speed of the stone. So we begin by recalling that if 𝑥 is a function for displacement at time 𝑡, then 𝑣 velocity is d𝑥 by d𝑡. It’s the change in displacement with respect to time. We need to be careful, though. There is a very subtle difference between speed and velocity. Velocity is a vector quantity. It has a direction, whereas speed doesn’t. Speed is magnitude of velocity.

We can represent the magnitude of the velocity or the absolute value of the velocity by using these bars. And so it follows that we’re going to begin by differentiating our expression for ℎ which we said is simply the displacement from the ground with respect to 𝑡. We’ll do this term by term. The derivative of 46.6𝑡 with respect to 𝑡 is just 46.6. Then, when we differentiate negative 4.9𝑡 squared, we multiply the term by the exponent and then reduce that exponent by one. So we get negative two times 4.9𝑡 or negative 9.8𝑡.

So we found an expression for dℎ by d𝑡. It’s 46.6 minus 9.8𝑡. And this, in fact, describes the velocity of the stone at time 𝑡 seconds. We want to work out the speed of the stone when it’s 22.5 meters high, in other words, when ℎ is equal to 22.5. So what we’re going to do is calculate the value of 𝑡 when ℎ is 22.5. This will tell us the time or times that the stone reaches this height. And then we can substitute that value of 𝑡 into our function for velocity. Letting ℎ be equal to 22.5, and we get a quadratic equation. It’s 22.5 equals 46.6𝑡 minus 4.9𝑡 squared.

Now, we have a number of techniques for solving quadratic equations, but each of those techniques only works if our equation itself is equal to zero. So we add 4.9𝑡 squared to both sides and subtract 46.6𝑡. So we get the equation 4.9𝑡 squared minus 46.6𝑡 plus 22.5 equals zero. We could use the quadratic formula completing the square or the quadratic equations solver on our calculator. When we do, we get 𝑡 equals nine and 𝑡 equals 0.51 and so on. So this tells us that the height of the stone is 22.5 meters at two different times. It’s when 𝑡 is equal to nine seconds, but also when 𝑡 is equal to 0.51 seconds.

Now, in fact, this makes a lot of sense, given the trajectory of the stone. The stone is projected upwards, but then it will come back down again. And so presumably, there will be two points at which the height of the stone is 22.5 meters. It will hit this height on its way up but then on its way down again. We need to double-check what happens with the velocity at both of these times. Let’s start by calculating the velocity of the stone when 𝑡 is 0.51 and so on. We’ll let 𝑡 be equal to this value. So velocity is 46.6 minus 9.8 times 0.51. That gives us a velocity of 41.6 meters per second.

And now, let’s repeat that process when 𝑡 is equal to nine. We get 46.6 minus 9.8 times nine, which is negative 41.6 meters per second. So the velocity on the way up is 41.6 meters per second and on the way down it’s the negative version of this. Now, it’s only negative because it’s traveling in the opposite direction. We said the speed is the magnitude of velocity or the absolute value. And so despite the fact that the stone is traveling in a different direction when 𝑡 is nine and when 𝑡 is 0.51, its speed is simply 41.6 meters per second. The speed of the stone is 41.6 meters per second.

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