A ball is thrown upward with an initial velocity of 13 m/s. Using the approximate value of

g = 10 m/s2, what are the magnitude and direction of the velocity: (a) 1 second after it is

thrown? (b) 2 seconds after it is thrown?

Respuesta :

Answer:

v=v0 - gt

Explanation:

The equation for velocity is

v=v0 - gt

where v0=14m/s, g=10m/s^2.

in 1 second:

v=14-10=4m/s

it is positive so direction is upwards

in 2 seconds:

v=14-20=-6m/s

it is negative so direction is downwards

The magnitude and direction of the velocity after 1 second and 2 seconds will be 3 m/sec and -7 m/sec,

What is velocity?

The change of distance with respect to time is defined as speed. Speed is a scalar quantity. It is a time-based component. Its unit is m/sec.

The given data in the problem is;

u is the initial velocity of fall = 13 m/sec

h is the distance of fall =  m

g is the acceleration of free fall = 9.81 m/sec²

v₁ is the velocity after 1 second =?

v₂ is the velocity after 2 seconds=?

According to Newton's second equation of motion,

v =u+at

a)

The magnitude and direction of the velocity after 1 second will be;

v =u+at

v=13+(-10)1

v=3 m/sec

+ ve shows the motion of the body is downward.

b)

The magnitude and direction of the velocity after 2 seconds will be;

v =u+at

v=13+(-10)2

v= -7 m/sec

- ve shows the motion of the body is upward.

Hence the magnitude and direction of the velocity after 1 second and 2 seconds will be 3 m/sec and -7 m/sec,

To learn more about the velocity, refer to the link: https://brainly.com/question/862972.

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