To determine the velocity and acceleration of the baseball at the top of its trajectory, we need to consider the motion of the baseball in a projectile motion.
Projectile motion can be analyzed independently in the horizontal (x) and vertical (y) directions because they are orthogonal to each other.
(a) Velocity at the top of the trajectory: At the top of the trajectory, the vertical velocity component becomes zero while the horizontal component remains constant. The vertical component of the velocity changes due to the acceleration due to gravity.
Given the initial velocity of the baseball as (10î + 15ĵ) m/s, we can break it down into its x and y components: Initial horizontal velocity (Vx) = 10 m/s Initial vertical velocity (Vy) = 15 m/s
At the top of the trajectory, the vertical velocity becomes zero, so Vy = 0. The horizontal velocity remains constant, so Vx remains unchanged as well.
Therefore, the velocity of the baseball at the top of its trajectory is (10î + 0ĵ) m/s or simply 10î m/s in the horizontal direction.
(b) Acceleration at the top of the trajectory: Neglecting air resistance, the only force acting on the baseball is the gravitational force, which causes a constant acceleration due to gravity in the vertical direction.
The acceleration due to gravity is always directed downward and has a magnitude of approximately 9.8 m/s^2 (assuming no significant variations in the gravitational field).
Therefore, at the top of the trajectory, the acceleration of the baseball is (0î - 9.8ĵ) m/s^2 or simply -9.8ĵ m/s^2 in the vertical direction. The negative sign indicates that the acceleration due to gravity is directed opposite to the positive y-axis.