A ball is shot from the ground into the air. At a height of 9.1 m, its velocity is v⃗=(7.6 î+6.

A ball is shot from the ground into the air. At a height of...

A ball is shot from the ground into the air. At a height of $9.1 \mathrm{~m}$, its velocity is $\vec{v}=(7.6 \hat{\mathrm{i}}+6.1 \hat{\mathrm{j}}) \mathrm{m} / \mathrm{s}$, with $\hat{\mathrm{i}}$ horizontal and $\hat{\mathrm{j}}$ upward. (a) To what maximum height does the ball rise? (b) What total horizontal distance does the ball travel? What are the (c) magnitude and (d) angle (below the horizontal) of the ball's velocity just before it hits the ground?

A ball is shot from the ground into the air. At a height of $9.1 \mathrm{~m}$, its velocity is $\vec{v}=(7.6 \hat{\mathrm{i}}+6.1 \hat{\mathrm{j}}) \mathrm{m} / \mathrm{s}$, with $\hat{\mathrm{i}}$ horizontal and $\hat{\mathrm{j}}$
upward. (a) To what maximum height does the ball rise? (b) What total horizontal distance does the ball travel? What are the
(c) magnitude and (d) angle (below the horizontal) of the ball's velocity just before it hits the ground?

Key Concepts

Vector Decomposition

Vector decomposition involves breaking down the projectile’s velocity into horizontal and vertical components. This is crucial for solving projectile problems because it allows each component to be analyzed independently using appropriate kinematic equations.

Horizontal Range

The horizontal range is the total horizontal distance traveled by the projectile. It is computed by multiplying the constant horizontal velocity by the total time of flight, which must be calculated based on the vertical motion under gravity.

Maximum Height

Maximum height in projectile motion is reached when the vertical component of velocity becomes zero. It can be determined using kinematic formulas that relate the initial vertical velocity, acceleration due to gravity, and vertical displacement, thereby identifying the peak point of the projectile’s trajectory.

Projectile Motion

Projectile motion refers to the motion of an object that is launched into the air and moves under the influence of gravity alone, neglecting air resistance. This concept highlights the independence of the horizontal and vertical components of motion, where gravity only affects the vertical motion and the horizontal motion remains at a constant velocity.

Kinematics Equations

Kinematics equations describe the motion of objects under constant acceleration. In the context of projectile motion, these equations are applied separately to horizontal displacement (constant velocity) and vertical displacement (accelerated motion due to gravity), enabling calculations such as maximum height, time of flight, and range.

Transcript

00:01 This problem is about projectile motion and in part a we want to find the maximum height of our projector and to do that we can split its vertical motion first in two parts because we're talking about height so we want to talk about the vertical displacement.

00:18 First we know it traveled 9 .1 meters but then it kept traveling until it stopped and then of course it went back down.

00:32 But for now let's talk about the heights.

00:35 And then we'll have this second height.

00:38 Now we want to find the total height by adding days to the first we need to find the second height.

00:45 So we can use the following equation, which tells us that velocity final squared is equal to the initial velocity squared plus two times the acceleration times the displacement.

01:00 We know that the final velocity is going to be zero and the initial velocity is going to be for our purposes at this point and we were told that it is 6 .1 meters per second so you use 6 .1 meters per second squared plus well it's actually going to be negative because the ball is de -accelerating so we have a negative acceleration and it is accelerating due to gravity so we have g and our displacement is height 2 that that gives us that height 2 is going to be equal to 6 .1 squared divided by two times gravity.

01:48 And taking gravity to be 9 .8, we get that this is roughly 1 .9 meters.

01:58 Therefore, our total height is going to be 1 .9 plus 9 .1.

02:05 And that is going to give us 10 meters in total.

02:11 Oh, i'm sorry.

02:15 That is 11 meters.

02:19 That is the maximum height.

02:21 Now let's move on to part b.

02:26 In part b, we want to find the horizontal distance.

02:30 And we know that in projectile motion, the horizontal velocity remains constant, given that we can neglect air resistance.

02:39 So we are going to say that our displacement in the x -axis is going to be equal to 7 .6 meters per second times the total time of the motion and we don't know the total time but we can find that really easily because we know the height and we know that it had to fall back down so we're going to use this equation we're going to say that time is equal to the square root of two times total height divided by gravity of course this is this is this time is not going to represent the time for the entire motion.

03:33 This equation just represents the time it took for the ball to fall back down.

03:38 So our total time is going to be this multiplied by two...

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