Two stones are thrown vertically upward from the ground, one...

Two stones are thrown vertically upward from the ground, one with three times the initial speed of the other. (a) If the faster stone takes 10 s to return to the ground, how long will it take the slower stone to return? (b) If the slower stone reaches a maximum height of $H$, how high (in terms of $H$) will the faster stone go? Assume free fall.

Key Concepts

Time of Flight

Time of flight is the total duration a projectile spends in the air before returning to its initial vertical position. For an object projected vertically upward under gravity, the time of flight is directly related to both the initial velocity and the acceleration due to gravity. Understanding this concept is crucial for comparing the flight durations of projectiles thrown with different speeds.

Maximum Height

Maximum height is the highest vertical position reached by a projectile relative to its starting point. It is determined by the initial speed and the constant acceleration due to gravity, with the relationship being quadratic in the initial velocity. This concept is essential for analyzing and comparing the peak altitudes achieved by objects launched with different initial speeds.

Uniformly Accelerated Motion

This concept refers to motion in which the acceleration remains constant over time. In the context of projectile motion, the only acceleration acting on the object (ignoring air resistance) is due to gravity, which remains constant near the Earth’s surface. This principle allows us to use the standard kinematic equations to analyze the upward and downward motions of the projectile.

Kinematic Equations

These are mathematical formulas that relate displacement, initial velocity, time, acceleration, and final velocity in situations involving uniform acceleration. They provide the tools necessary to describe and predict the behavior of projectiles and are especially useful when determining quantities like maximum height and time of flight in vertical motion.

Transcript

00:01 Part a of this problem is asking what the total elapse time of the slower rock will be on its entire journey.

00:07 Before we're able to calculate that, we need to figure out what the initial velocity of the slower rock is along its journey.

00:14 And so we're given this information.

00:16 The acceleration is negative 9 .8.

00:18 The initial position is zero.

00:20 The initial speed of or the initial velocity of the slow rock is just the initial velocity of the slow rock.

00:27 And then of the large rock is three times a slow rock.

00:29 And then the time elaps for the faster rock is 10 seconds.

00:33 So since the time elaps for the total trip is 10 seconds, we know that it takes 5 seconds for the fast rock to reach peak height.

00:55 And then once it does reach peak height, our velocity will be zero.

01:01 And this is velocity of 2, which i call the faster rock.

01:06 Now, in order to figure out the initial velocity, i'm going to use this kinemak equation.

01:12 V is equal to v0 plus acceleration times time.

01:17 V at the peak is zero.

01:19 Our initial velocity is, this is of the fast rock, by the way, so v02, which is 3 v .0 .1 plus the acceleration negative 9 .8 times the time.

01:34 And this is the time from the start until it reaches peak height, which is 5 seconds.

01:40 We can solve this for the initial velocity of the slower rock.

01:45 We get 16 .33 meters per second.

01:52 Now that we have this, we can figure out the total time for the smaller rock.

01:59 So if v01 is equal to 16 .33 meters per second, we can use the same kinematic equation.

02:11 V is equal to v0 plus at.

02:16 The difference is before i applied it to the faster rock in order to find this, but now i'm going to apply it to the slower rock, now that we know this.

02:26 And because we're playing it to the slower rock, i got to remember to put that subscript there, the one there.

02:32 And so we're just going to go up to the peak.

02:35 So the final velocity is zero.

02:38 This is 16 .33.

02:41 Acceleration is still negative 9 .8.

02:44 And then t.

02:46 Solving, we get t is equal to 1 .67 seconds...