The quantity that has a magnitude of zero when the ball is at the highest point in its trajectory is
the vertical velocity.
In fact, the motion of the ball consists of two separate motions:
- the horizontal motion, on the x-axis, which is a uniform motion with constant velocity

, where

- the vertical motion, on the y-axis, which is a uniformly accelerated motion with constant acceleration

directed downwards, and with initial velocity

. Due to the presence of the acceleration g on the vertical direction (pointing in the opposite direction of the initial vertical velocity), the vertical velocity of the ball decreases as it goes higher, up to a point where it becomes zero and it reverses its direction: when the vertical velocity becomes zero, the ball has reached its maximum height.
The motion of the ball is a composition of two motions:
- on the x (horizontal) axis, it is a uniform motion with initial velocity

- on the y (vertical) axis, it is a uniformly accelerated motion with acceleration
(a) to solve this part, we just analyze the motion on the vertical axis. The law of motion here is

By requiring y(t)=0, we find the time t at which the ball reaches the floor:


(b) for this part, we can analyze only the motion on the horizontal axis. To find how far the ball will land, we must calculate the distance covered on the x-axis, x(t), when the ball reaches the ground (so, after a time t=0.64 s):
Answer:
1.) A simple harmonic oscillator has an amplitude of 3.50 cm and a maximum speed of 26.0 cm/s. What is its speed when the displacement is 1.75 cm? 2.) Both pendulum A and B are 3.0 m long. The period of A is T. Pendulum A is twice as heavy as pendulum B. What is the period of B? 3.) The time for one cycle of a periodic process is called the _ ? 4.) In simple harmonic motion, the acceleration is proportional to? 5.) The position of a mass that is oscillating on a spring is given by x= (18.3 cm) cos [(2.35 s-1)t]. What is the frequency of this motion?
Explanation:
B. Chemical reactions must overcome the strong nuclear force