An object is dropped from a height of 75.0 m above ground level. (a) Determine the distance trav

step 1 So here it's a bit easier to choose downwards to just be positive. And so we can say that for, i

An object is dropped from a height of 75.0 m above ground...

An object is dropped from a height of $75.0 \mathrm{m}$ above ground level. (a) Determine the distance traveled during the first second. (b) Determine the final velocity at which the object hits the ground. (c) Determine the distance traveled during the last second of motion before hitting the ground.

Key Concepts

Uniformly Accelerated Motion

This concept refers to motion in which an object’s acceleration remains constant throughout the motion. In the context of free fall, the acceleration is due to gravity and is considered uniform (typically taken as 9.8 m/s² or 10 m/s² in approximate calculations), which allows the use of kinematic equations to predict velocity and displacement at various time intervals.

Acceleration due to Gravity

Acceleration due to gravity is the constant rate at which objects accelerate when falling under the influence of Earth’s gravitational field. This value, approximately 9.8 m/s², is a key parameter in analyzing free fall motion and is used in kinematic equations to calculate changes in velocity and displacement over time.

Kinematic Equations

Kinematic equations describe the relationships among displacement, velocity, acceleration, and time for an object moving with constant acceleration. These equations are essential tools in solving problems involving free fall, such as determining the distance traveled in a specific time interval, the final velocity of an object, or the total time of flight from a given height.

Time Interval Analysis

This concept involves breaking down the overall motion into specific time intervals to analyze aspects such as the distance covered or change in velocity during those intervals. In free fall problems, analyzing intervals like the first second or the last second of motion helps in applying kinematic equations more precisely to obtain insights about the object's behavior over different portions of its journey.

Transcript

00:01 So here it's a bit easier to choose downwards to just be positive.

00:04 And so we can say that for, if we say this, we can say that y final is equaling 75 .0 meters.

00:15 And we can choose the place with which the object is dropped to y initial.

00:20 This would equal zero meters.

00:21 And so for part a, when we say that y final equals y initial plus vy initial t plus one half gt squared, well, in the first second, y initial is equaling 0, vy initial is equaling 0.

00:39 So both of these are 0.

00:41 And so y final would be equaling 1 1⁄2 meters per second squared times 1 .00 seconds quantity squared.

00:52 This is simply equaling essentially 4 .90 meters.

00:58 So it's traveled 4 .90 meters.

01:01 In the first second.

01:04 And then it's saying, essentially, what's the speed with which it hits the ground? if, again, we're choosing downwards to be positive, we expect the positive velocity.

01:15 We should take note of this downwards is positive.

01:26 And so we can say that v .y final squared equals v .y initial squared plus two times the acceleration due to gravity times delta y.

01:36 And in this case, we know that initially we don't have any velocity in the y direction.

01:42 So vy final is equaling the square root of 2g delta y, and we're traveling 75 meters.

01:49 So this is equalling the square root of 2 times 9 .8 meters per second squared, multiplied by 75 .0 meters.

02:00 Now, we can say that vy final is going to be equal in positive 38 .3 meters per second.

02:09 And all we have to do is just note this is downwards.

02:13 So it depends on the situation, but sometimes choosing downwards to be positive makes calculating everything much, much easier.

02:23 And then we have to, for part c, we want to determine the distance travel during the last second of motion before hitting...

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