Question

A heavy ball is thrown zero-launch from a tower with an initial velocity of 35 m/s. (use g = 10 m/s² down for gravity). After 4.0 s, what is the magnitude of the x and y component of the velocity?

Option 1: 35 m/s and 0 m/s
Option 2: 0 m/s and 35 m/s
Option 3: 24 m/s and 20 m/s
Option 4: 20 m/s and 24 m/s

Answer 1

The magnitude of the x component of the velocity is 35 m/s and the magnitude of the y component of the velocity is 40 m/s.

Step-by-step explanation:

The magnitude of the x component of the velocity remains constant throughout the motion as there is no horizontal acceleration. Therefore, the magnitude of the x component of the velocity is equal to the initial velocity, which is 35 m/s.

The vertical component of the velocity changes due to the effect of gravity. After 4.0 s, the vertical component of the velocity can be found using the equation:

vy = v0y + gt

where vy is the vertical component of the velocity, v0y is the initial vertical component of the velocity (which is 0 m/s as the ball is thrown horizontally), g is the acceleration due to gravity (-10 m/s²), and t is the time (4.0 s).

By substituting the values, we get:

vy = 0 + (-10)(4.0)

vy = -40 m/s

Therefore, the magnitude of the y component of the velocity is 40 m/s.