Question

A baseball player at bat hits a ball to the outfield. The baseball is hit when it is at a height of 1.60 m above the home plate. The ball’s velocity the instant it is hit is 34.00 m/s at an angle of exactly 54.00^o above the horizontal. The ball lands safely in the outfield. Assume that the entire baseball ground is horizontal.

(a) What are the vertical and horizontal components of the ball’s acceleration while it is in the air?

(b) What are the vertical and horizontal components of the baseball’s speed the instant after it was hit?

(c ) What is the maximum height of the baseball above the ground?

(d) What is the Range of the baseball?

(e) What is the velocity (vector) of the ball the instant before it hits the ground?

Answer 1

PART A:

There is no force acting on the ball in the horizontal direction. So, the acceleration in the horizontal direction is zero. i.e

There is the gravitational force acting on the ball in the downward vertical direction. So, the acceleration in the vertical direction is due to gravitation. i.e

PART B:

The vertical component of the velocity after the ball was hit is

The horizontal component of the ball is

PART C:

At the maximum height, the vertical velocity of the ball is zero. i.e.

So, the maximum height is 38.6 m.

PART D:

The range of the projectile is given by

So, the range of this projectile is 112.2 m.

PART E:

When it hits the ground, the vertical component of the velocity is in opposite direction to the initial vertical velocity with the same magnitude while the horizontal component remains constant throughout the projectile. So, the velocity vector is