Question Video: Finding the Maximum Height Attained by an Object Projected Vertically Upwards Mathematics

A particle is projected vertically upwards at 7 m/s from a point 38.7 m above the ground. Find the maximum height that the particle can reach. Consider the acceleration due to gravity to be 9.8 m/s².


Video Transcript

A particle is projected vertically upward at seven meters per second from a point 38.7 meters above the ground. Find the maximum height the particle can reach. Consider the acceleration due to gravity to be 9.8 meters per second squared.

Alright, so if we say that this is ground level, then if we go 38.7 meters up from there, we have our particle where it is projected with an upward velocity of seven meters per second. This tells us that our particle will continue moving upward. But we know that under the influence of gravity, it will slow down more and more until eventually it comes to rest. At this point, it has reached its maximum height. We’ll call it ℎ sub max. This is the height we want to solve for, and we can see it’s equal to 38.7 meters plus this height here. We’ll call this height ℎ. And since as our particle moves across this height it’s accelerating uniformly under the influence of gravity, we can use an equation of motion to solve for ℎ. Specifically, we’ll use the relationship that an object’s final velocity squared is equal to its initial velocity squared plus two times its acceleration times its displacement.

Applying this relationship to our situation, we’ll say 𝑣 sub 𝑓 squared equals 𝑣 sub 𝑖 squared plus two times 𝑔, that’s our particle’s acceleration, multiplied by ℎ. Going back over to our sketch, if we say that our particle’s position here is its initial position, where it had an initial velocity 𝑣 sub 𝑖, and our particle’s position here at its maximum height is its final position, then we can say that its final velocity 𝑣 sub 𝑓 is zero, meaning that this is then the equation we have to solve for the height ℎ.

At this point, we can set up a sign convention where we say that motion upward is positive and motion downward is therefore negative. This is useful to us because the acceleration due to gravity is downward, while our initial velocity, what we’ve called 𝑣 sub 𝑖, is upward. And so leaving off the units, this means that we would use a value of positive seven for 𝑣 sub 𝑖 and negative 9.8 for 𝑔. Substituting in these values, if we then subtract seven squared from both sides of this equation, we have that negative seven squared is equal to two times negative 9.8 times ℎ. And both negative signs drop out. And if we then divide both sides of the equation by two times 9.8, we find that ℎ is equal to 49, that’s seven squared, divided by two times 9.8. That’s equal to 2.5. And we’ll include the units of meters.

Recall though that this isn’t our answer because ℎ is just one part of ℎ max. ℎ max is equal to 38.7 meters plus ℎ or 38.7 meters plus 2.5 meters. And adding these together, we get 41.2 meters. This is the maximum height our particle can reach.

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