a) When the ball reaches the top of its trajectory, its velocity will be solely in the y-direction (upwards).
Since at the top of its trajectory, the ball momentarily stops moving vertically before it starts to fall back down, the velocity at the top of its trajectory will be 0 m/s in the y-direction.
Therefore, the velocity at the top of its trajectory is ⃗ ( ̂) m/s.
b) At the top of its trajectory, the acceleration of the ball will be solely due to gravity, pointing downwards.
The acceleration due to gravity is 9.8 m/s^2 in the negative y-direction.
Therefore, the acceleration at the top of its trajectory is ⃗ (− ̂) 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