A rock thrown vertically upward from the surface of the moon...

A rock thrown vertically upward from the surface of the moon at a velocity of $24 \mathrm{m} / \mathrm{s}$ (about $86 \mathrm{km} / \mathrm{h}$ ) reaches a height of $s=24 t-0.8 t^{2} \mathrm{m}$ in $t \mathrm{s}$. a. Find the rock's velocity and acceleration at time $t .$ (The acceleration in this case is the acceleration of gravity on the moon.) b. How long does it take the rock to reach its highest point? c. How high does the rock go? d. How long does it take the rock to reach half its maximum height? e. How long is the rock aloft?

A rock thrown vertically upward from the surface of the moon at a velocity of $24 \mathrm{m} / \mathrm{s}$ (about $86 \mathrm{km} / \mathrm{h}$ ) reaches a height of $s=24 t-0.8 t^{2} \mathrm{m}$ in $t \mathrm{s}$.
a. Find the rock's velocity and acceleration at time $t .$ (The acceleration in this case is the acceleration of gravity on the moon.)
b. How long does it take the rock to reach its highest point?
c. How high does the rock go?
d. How long does it take the rock to reach half its maximum height?
e. How long is the rock aloft?

Key Concepts

Projectile Motion

Projectile motion involves the motion of an object subject to a constant acceleration, usually gravity, in one or more dimensions. It is fundamental in understanding how objects move when they are thrown, taking into account initial velocity, displacement, and the effects of gravitational force. For problems like these, the equations of motion govern both the upward and downward trajectories under uniform acceleration.

Differentiation in Kinematics

Differentiation is used in kinematics to move between displacement, velocity, and acceleration. By taking the first derivative of the displacement function with respect to time, one obtains the velocity function, and the second derivative gives the acceleration function. This process allows the analysis of the speed and the rate of change of speed at any given moment in a projectile’s path.

Acceleration Due to Gravity

The acceleration due to gravity is a constant value that describes the rate at which objects accelerate when falling solely under gravity’s influence. On the Moon, this value is significantly lower than on Earth, affecting the projectile’s motion. Understanding this concept is crucial for calculating the projectile’s acceleration and predicting its overall motion.

Maximum Height

The maximum height reached by a projectile is achieved when its velocity becomes zero, marking the end of the upward motion. By setting the velocity function to zero, one can solve for the time at which this occurs and then substitute this time back into the displacement equation to determine the highest point reached.

Time of Flight

The time of flight is the total duration that a projectile remains in the air from launch to when it returns to its original launch level. It is determined by analyzing the displacement equation and solving for the time when the displacement equals the initial position, taking into account both the ascent and descent periods.

Quadratic Equations in Kinematics

Many kinematics problems, especially those involving projectile motion with constant acceleration, lead to quadratic equations. Solving these equations is essential for finding specific time intervals in the motion, such as the time to reach a particular height or the overall duration of the flight. Techniques for solving quadratic equations, such as the quadratic formula, are therefore a key part of the analysis.

Transcript

00:01 The rock thrown vertically upward from the surface of the moon at a velocity of 24 meters per second reaches a height of s equals 24 t minus 0 .8 t squared meters in t seconds.

00:14 A, find the rock's velocity and acceleration at time t.

00:19 Okay, so the velocity and acceleration are based on derivatives of the position vector.

00:30 So the height above the ground is s equals 24t minus 0 .8 t squared meters.

00:42 So dsdt, which will give the magnitude of the velocity, is 24 minus 1 .6t.

00:53 So in other words, the velocity vector as a function of t, is 24 minus 1 .6t.

01:02 6 t meters per second and this is going to be in the z direction because the velocity is either always up or always down because we're just considering up -down one -dimensional motion and then the acceleration is the second derivative of t which of course is just the derivative of the derivative so the derivative of v the acceleration and the derivative of v and the derivative of v is negative 1 .6 meters per second squared.

01:51 Again, in the z direction.

01:54 The minus 1 .6 just comes from taking this derivative and that term right there.

02:00 Okay, so that's our acceleration.

02:04 Second part of the question, b, how long does it take the rock to reach its highest point? so the highest point occurs when the velocity is zero, right? it's when the trajectory is going up and then it reaches a maximum point and it momentarily stops.

02:24 At this point, dsdt is zero.

02:28 So we want to find a dsdt at some maximum, at some time that corresponds to the maximum height, where the speed is zero.

02:42 So this is going to be 24 minus 1 .6 t sub m.

02:48 T sub m is this time that we want.

02:51 So t sub m solving for that because we can just throw it on this side of the equation, divide by 1 .6.

02:58 We'll get 24 over 1 .6 seconds.

03:03 So t sub m is, therefore, 24 over 1 .6 is 15, 15 seconds.

03:21 And that's how long it reaches the highest point.

03:26 And, okay, how high does the route go? okay, how high does it go when it goes to the highest point, which we know is at 15 seconds? let's just plug in 15 for, or let's plug in t -max into s.

03:41 And 24 times 15, minus 0 .8 times 15 squared.

03:50 Okay, let's plug that in.

03:53 24 times 15 minus 0 .8 times 15 times 15 times 15 squared.

04:06 180.

04:08 So s of t sub m is 180 meters.

04:15 That's how high the thing goes...

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