A tennis ball is served from the back line of the court such...

A tennis ball is served from the back line of the court such that it leaves the racket $2.4 \mathrm{m}$ above the ground in a horizontal direction at a speed of $22.3 \mathrm{m} / \mathrm{s}(50 \mathrm{mi} / \mathrm{h}) .$ (a) Will the ball cross a $0.91-\mathrm{m}$ -high net $11.9 \mathrm{m}$ in front of the server? (b) Will the ball land in the service court, which is within $6.4 \mathrm{m}$ of the net on the other side of the net?

Key Concepts

Projectile Motion

Projectile motion is the study of objects that are launched into the air and move under the influence of gravity alone (assuming air resistance is negligible). It combines both horizontal motion with constant velocity and vertical motion with constant acceleration due to gravity. This concept forms the basis for analyzing the object’s trajectory, duration of flight, and displacement.

Separation of Motion Components

In projectile motion, the horizontal and vertical components of the velocity are treated independently. The horizontal motion proceeds at a constant velocity because there is no acceleration in that direction (under ideal conditions), while the vertical motion is influenced by gravity. This separation simplifies the analysis and calculation of the object's path.

Kinematic Equations

Kinematic equations describe the relationships among displacement, speed, acceleration, and time for objects in motion. In the context of projectile motion, different equations are applied to the horizontal and vertical components to predict positions, compute the time of flight, and determine whether the projectile will clear obstacles or land within a certain area.

Time of Flight and Horizontal Range

The time of flight refers to the duration an object remains in motion before landing, and the horizontal range is the total distance traveled in the horizontal direction. These calculations are essential when determining if a projectile will reach a specific target area or clear a particular barrier.

Trajectory Analysis

Trajectory analysis involves determining the curved path followed by a projectile by combining its horizontal and vertical motions. By analyzing the trajectory using equations from physics, one can calculate heights at given horizontal distances, which is crucial for assessing whether the projectile will pass over obstacles such as nets or land within designated boundaries.

Transcript

00:02 Over there first of all we need to arrange the data which is even so basically the initial of the first one is given that is 22 .3 and i directs in a meter per second and and this means in the direction of x over there and y i 1 over there is nothing for 2 .4 meter actually the height in the vertical direction and x i i have 1 is 0 meter that means at the horizontal level and actually or and the xf1 is 11 .9 meter this is at the origin level of this one and finally it is on the 11 .9 meter you know ahead of the origin and h net or h net over there are final over there is 0 .91 meter over there which means that initial velocity over there is 0 meter per second and and that final velocity is 2 .3 meter this means over there so basically for the first part of this question to find if the ball will cross the height of which is given 0 .91 meter net or not we need to find the height of the wall when it travel at horizontal distance of 11 .9mere obviously this is a pretty easy question as well so we need to find the time it takes to travel this horizontal distance by using kinematic formula displacement.

01:41 So basically we know that s equals to ut plus half a t square or you can say that x, f or final minus of initial, is equal to v ix, vut plus half into axt square over there.

01:59 And that means delta x over there will be nothing but b ix.

02:03 T plus half a x t square not our general acceleration is zero because we suppose that there is no drag force this exerted on the ball during the movement in the direction so this this term over there will be zero so we can clearly see that delta x over there will be nothing but v i x entity so t over there would be nothing but delta x upon v i x and that it means delta x over there is given which is 11 .9 and initial velocity is also given 22 .3.

02:37 So time t over there comes up to be 0 .534 second.

02:46 And so actually the figure over there.

02:50 So basically actually the figure over there is look like this one.

02:52 Like say this is initially height of height is 2 .4 meter and when it crosses to 11 .9 meter should be height of an, it should be height of actually 0 .4 of 9 1 meter actually so we need to determine 9.

03:10 So we need to determine over there.

03:13 This height is actually that one and the ball is over there...

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