m/s (b) Find the time it takes the ball to reach the wall. (c) Find the velocity components of the ball when it reaches the wall. x-component m/s y-component m/s Find the speed of the ball when it reaches the wall. m/s
Displacement, Velocity and Acceleration
In classical mechanics, kinematics deals with the motion of a particle. It deals only with the position, velocity, acceleration, and displacement of a particle. It has no concern about the source of motion.
Linear Displacement
The term "displacement" refers to when something shifts away from its original "location," and "linear" refers to a straight line. As a result, “Linear Displacement” can be described as the movement of an object in a straight line along a single axis, for example, from side to side or up and down. Non-contact sensors such as LVDTs and other linear location sensors can calculate linear displacement. Non-contact sensors such as LVDTs and other linear location sensors can calculate linear displacement. Linear displacement is usually measured in millimeters or inches and may be positive or negative.
![### Projectile Motion Problem
#### Problem Statement:
A home run is hit in such a way that the baseball just clears a wall 12.0 m high, located 126 m from home plate. The ball is hit at an angle of 37.0° to the horizontal, and air resistance is negligible. (Assume that the ball is hit at a height of 1.0 m above the ground.)
#### Questions:
- **(a)** Find the initial speed of the ball.
- [Answer] _______ m/s
- **(b)** Find the time it takes the ball to reach the wall.
- [Answer] _______ s
- **(c)** Find the velocity components of the ball when it reaches the wall.
- **x-component:** [Answer] _______ m/s
- **y-component:** [Answer] _______ m/s
- Find the speed of the ball when it reaches the wall.
- [Answer] _______ m/s
#### Submission:
- Click the "Submit Answer" button when you are ready to submit your answers.](/v2/_next/image?url=https%3A%2F%2Fcontent.bartleby.com%2Fqna-images%2Fquestion%2F8fa35088-fa5f-472d-877a-bca1a109875f%2F16a5f810-ac40-4fe6-8a3c-22e7800ebb86%2F3e6hjyf_processed.png&w=3840&q=75)
Max height attained by the ball =12 m at an angle of 37 degree
Ma x height= v2sin2¶/2g
Where ¶= angle of elevation
So, max height=v2(sin37) 2/2g
√(12*2*9.8/(sin37) 2)=v
v=25.48m/sec
Time taken by the ball to reach the ground
Time from ground to the max height
T1=vsin¶/g
T2=Time from max height to 1 m above ground
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