a) At the top of the trajectory, the velocity of the ball will be purely in the \(y\) direction, since the ball reaches its maximum height. The velocity in the \(x\) direction will be zero. Therefore, the velocity of the ball at the top of its trajectory will be \(0 ̂+v_{y} ̂\), where \(v_{y}\) is the \(y\) component of the initial velocity.
Given that the initial velocity is \((2 ̂+3 ̂) \, \text{m/s}\), the \(y\) component of the initial velocity is \(3 \, \text{m/s}\). Thus, the velocity of the ball at the top of its trajectory is \(0 ̂+3 ̂ \, \text{m/s}\).
b) At the top of the trajectory, the acceleration of the ball is solely due to gravity, acting in the negative \(y\) direction. The magnitude of the acceleration due to gravity is approximately \(9.81 \, \text{m/s}^2\).
Therefore, the acceleration at the top of the trajectory is \((0 ̂-9.81 ̂) \, \text{m/s}^2\).
A ball is thrown with an initial velocity of( ̂+ ̂) m/s. When it reaches the top of its
trajectory, neglecting air resistance, what is its a) velocity? b) Acceleration?
1 answer
1 answer