You throw a baseball from the roof of a house to a friend on the ground. The ball has an
initial velocity of 12.0 m/s in the horizontal direction. After 1.00 s, how fast is the ball
moving in (a) the x direction and (b) the y direction. A) the velocity in the x-direction
doesn't change. Therefore, after 1.00 seconds we have
Please help!!

The velocity of the ball in the x-direction (horizontal direction) after 1 second = 12.0 m/s because there is no net acceleration in the horizontal direction

b) The velocity in the y-direction is given by the following kinematic equation;

v = u - g × t

Where;

v = The final vertical velocity of the ball after t seconds

u = The initial vertical velocity = 0

g = The acceleration due to gravity = 9.81 m/s²

t = The time taken in motion = 1.00 s

Substituting the values gives;

v = 0 - 9.81 m/s² × 1.00 s = -9.81 m/s

v = -9.81 m/s

The vertical velocity of the ball after 1.00 second = -9.81 m/s which is 9.81 m/s downwards.

onyebuchinnaji onyebuchinnaji · Answer 2 · 2021-12-03T13:27:32+01:00

(a) The final velocity of the ball after 1,0 second in y - direction is -9.8 m/s.

(b) The final velocity of the ball after 1,0 second in x - direction is 12 m/s.

The given parameters:

initial horizontal velocity of the ball, u = 12 m/s
time of motion, t = 1 s

The final velocity of the ball after 1,0 second in y - direction is calculated as follows;

[tex]v_f_y = v_0_y - gt\\\\v_f_y = 0 - (9.8\times 1)\\\\v_f_y = -9.8 \ m/s[/tex]

The final velocity of the ball after 1,0 second in x - direction is calculated as follows;

[tex]v_x_f = v_0_x \\\\v_x_f = 12 \ m/s[/tex]

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