You can shoot an arrow straight up so that it reaches the top of a 25-m -tall building. (a) How

You can shoot an arrow straight up so that it reaches the...

Key Concepts

Projectile Motion

Projectile motion is the study of objects that are launched into the air subject only to the acceleration of gravity. In these problems, the horizontal and vertical components of motion are analyzed separately, with the horizontal motion proceeding at constant velocity and the vertical motion under constant acceleration (gravity). This framework allows one to predict trajectories, determine distances, and calculate time of flight, regardless of the angle of launch.

Conservation of Energy

The conservation of energy principle states that energy cannot be created or destroyed, only transformed from one form to another. In the context of projectile problems, this concept is used to equate the initial kinetic energy of the projectile with the gravitational potential energy at its highest point. This relationship provides a method to determine the initial speed or energy imparted to the projectile when launched.

Kinematic Equations

Kinematic equations describe the motion of objects under constant acceleration, such as gravity. These equations relate displacement, initial velocity, final velocity, acceleration, and time, and are essential tools for solving projectile motion problems. They allow one to solve for various unknowns in the projectile's trajectory when given initial conditions and specific parameters of motion.

Vector Decomposition

Vector decomposition involves breaking a vector, such as the initial velocity of a projectile, into its horizontal and vertical components. This step is crucial in projectile motion analysis because it allows the use of kinematic equations independently in each direction. Understanding how to resolve an angled launch velocity into its components is fundamental in determining factors such as range, maximum height, and time of flight.

Idealized Assumptions

In many projectile motion problems, several idealizations are made to simplify the analysis. These include assuming a constant gravitational field (ignoring variations with altitude), neglecting air resistance, treating the projectile as a point mass, and considering a level launch and landing height when appropriate. These assumptions make the complex real-world behavior tractable with basic mathematical tools.

Transcript

00:01 In this question we have a we shoot ball or things over there which is 25 meters of the ground basically you can see that over there y final is nothing but 25 meter and v final over there is also 0 meter per second because at the maximum height for c2 and y initially is already zero you already know and this is the first scenario and for the second scenario we have y initial initial height over there is just 1 .45 meter and y final obviously will be 0 when the ball will grounded and x initial over there is 0 but v or i y initially zero over there so but this is actually thrown in the horizontal direction that's why and and when the arrow was a high 30 degree above the horizontal from the ground is this is the third scenario over there which is over there yf over there is 1 .45 meter and height of the target is just 1.

01:01 0 .45 meter as well and theta initial over there is 30 degree which is angle with the horizontal so basically for the first one we need to calculate the velocity so we already know that v squared equals to u s plus 2 as so v over there this is is 0 at the maximum height so 0 equals to u s square u s square and minus of minus will be obviously over there because g in the downward direction so minus of 2 g h and h over there is 25 meter so basically u over there that comes out to be root over two times of g which is 9 .8 into 25 so basically u over there that comes out to be 22 .1 meter per second and that will be plus minus as well obviously so we neglect the negative because it erode the find upward and initial velocity is positive so basically we don't tell about this one so this was the things over there the initial velocity is just 22 .1 per second and now we are going to solve first part of this problem that it's uh we will use the initial velocity which is found in the equation one and two so basically you're going to use that so y f minus of y i is nothing but b i y t plus half of gt square this is the at s square actually but in the place of a we can put g over there and y final over there is also zero v y initial velocity in the virus and a zero and so basically minus of y of y y i equals to nothing but minus of half g t square why is that why is that because g in the opposite direction so don't know direction that's why so this and this again get cancelled so t over the that comes up to be root over 2 y i upon g and two times of y i which is 1 .45 upon g is 9 .8 so time over there that comes out to be 0 .5 4 seconds so now we're going to calculate the horizontal range over there so for the horizontal range you already know that delta x or x would be nothing but really because an acceleration in the x is 0 so v i x into t so v i x is obviously we have calculated which is 22 .1 and time is 0 .54 so basically when delta x over there that comes out to be 119 meter okay now moving back to the next part of this problem in which let's suppose that you do length in suppose that your length is about 165 meter on your arm about 1 .45 meters from your feet.

03:41 So this means you will really lead arrow from the same height as the target.

03:45 Now, we need to find the same time it takes to reach out the same height again, which is mean delta y over there is zero.

03:53 So delta y is 0, delta y is equal to v iyy t plus plus half of gt square or radius square.

04:02 So delta y over there is zero...

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