You throw a ball toward a wall at speed 25.0 m / s and at angle θ0=40.0^∘ above the horizontal (

step 1 This is chapter 4 problem number 32. We're kicking the ball at a horizontal angle of 40 degrees

You throw a ball toward a wall at speed 25.0 m / s and at...

You throw a ball toward a wall at speed 25.0 $\mathrm{m} / \mathrm{s}$ and at angle $\theta_{0}=40.0^{\circ}$ above the horizontal (Fig. $4-35 )$ . The wall is distance $d=$ 22.0 $\mathrm{m}$ from the release point of the ball. (a) How far above the release point does the ball hit the wall? What are the (b) horizontal and (c) vertical components of its velocity as it hits the wall? (d) When it hits, has it passed the highest point on its trajectory?

You throw a ball toward a wall at speed 25.0 $\mathrm{m} / \mathrm{s}$ and at angle
$\theta_{0}=40.0^{\circ}$ above the horizontal
(Fig. $4-35 )$ . The wall is distance $d=$
22.0 $\mathrm{m}$ from the release point of the ball. (a) How far above the release
point does the ball hit the wall?
What are the (b) horizontal and (c) vertical components of its velocity as it hits the wall? (d) When
it hits, has it passed the highest point on its trajectory?

Key Concepts

Trajectory and Maximum Height Analysis

Understanding the shape of the projectile’s path, specifically the parabolic trajectory, helps in determining key features such as the highest point. Analyzing if the object has passed its maximum height at a given position involves comparing the time of flight to the time needed to reach the peak of the trajectory.

Time of Flight and Position Determination

Determining the time at which a projectile reaches a specific horizontal distance (such as where it hits a wall) is essential. Once the time is known, it can be used with kinematic equations to calculate other aspects of the motion, such as vertical displacement and the velocity components at that point.

Kinematic Equations

These equations relate displacement, velocity, acceleration, and time. In projectile motion, they are used separately for the horizontal and vertical components, with the horizontal motion having zero acceleration and the vertical motion influenced by the constant acceleration due to gravity.

Projectile Motion

This concept describes the motion of an object that is launched into the air and moves under the influence of gravity alone. The object’s trajectory is determined by both its initial velocity and the acceleration due to gravity, which only affects the vertical motion, while the horizontal motion remains constant (ignoring air resistance).

Decomposition of Velocity Vectors

Any velocity can be split into horizontal and vertical components using trigonometry. This decomposition is crucial in analyzing projectile motion since it allows solving for independent motions in the horizontal (constant velocity) and vertical (accelerated motion) directions.

Transcript

00:01 This is chapter 4 problem number 32.

00:03 We're kicking the ball at a horizontal angle of 40 degrees with an initial velocity of 25 meters per second against a wall that is 22 meters away from us.

00:24 And in part a, we're asked how far above the release point does the ball hit the wall? so it's going to eventually hit the bowl at some point, right? and let's call this height as h.

00:37 And we're exactly asked what h is in this case.

00:42 Now, let's look at the displacement in the y direction equals the initial y, the velocity in the initial y direction times time minus 1 .5 gt squared.

00:54 So the alpha y stands for y minus y final minus y initial.

01:00 Initial happens to be zero.

01:02 Then final is h.

01:03 And then what we're looking for is basically b0y, the y component of the velocity times t minus one -half gt square.

01:13 However, we don't know what time is.

01:16 It's not given to us in a problem.

01:19 However, what we know is that the distance d is going to be traveled with what speed in the extraction, right? so the time would be equal to the distance traveled in the extraction times.

01:32 The velocity in initial velocity in the extraction because this speed does not change in the horizontal direction due to not having an acceleration in the extraction so the over v .0 cosine theta then would give us the time so 22 minutes divided 22 meters pardon me divided by b not is 25 meters per second times cosine of 40 degrees, then the time can be found to be 1 .149 seconds.

02:08 Now, all we need to do is to plug it in the formula for height, d .0 is 25 meters per second, times side of 40 degrees times 1 .149 seconds minus 1 .5, 9 .8 meters per second square, times 1 .149 second square...

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