To determine the velocity and acceleration of the baseball when it reaches the top of its trajectory, we need to consider the kinematic equations for projectile motion.
Given: Initial velocity (u) = 10î + 15ĵ m/s (where î and ĵ are the unit vectors in the x and y directions, respectively)
At the top of its trajectory, the vertical velocity component becomes zero since the baseball momentarily stops before falling back down due to gravity.
(a) Velocity at the top of the trajectory: The horizontal component of velocity (in the x-direction) remains constant throughout the motion since there are no external forces acting in that direction. Therefore, the horizontal velocity component (Vx) remains the same as the initial velocity:
Vx = 10î m/s
For the vertical component (in the y-direction), at the top of the trajectory, the velocity becomes zero (Vy = 0) momentarily before the ball starts descending. This happens because the gravitational force acts against the initial upward velocity.
Velocity at the top of the trajectory: V = Vx + Vy = 10î + 0ĵ = 10î m/s
(b) Acceleration at the top of the trajectory: The acceleration due to gravity (g) acts vertically downward and remains constant throughout the motion. Therefore, the acceleration can be written as:
a = 0î - gĵ
The negative sign indicates that the acceleration due to gravity is in the opposite direction of the positive y-axis.
Acceleration at the top of the trajectory: a = -gĵ
In summary: (a) The velocity at the top of the trajectory is 10î m/s. (b) The acceleration at the top of the trajectory is -gĵ, where g represents the acceleration due to gravity.