a pitcher throws a softball horizontally from a height of 1.25 m and notes that it strikes the ground a distance of 15 m from the point where it was thrown. how long did it take the ball to reach the ground? what was the velocity at which the ball was thrown? what was the vertical component of the ball's velocity when it hit the ground?

Show steps please!

To find the time it took for the ball to reach the ground, we can use the formula for vertical motion:

h = (1/2) * g * t^2

Where:
h is the initial height (1.25 m in this case)
g is the acceleration due to gravity (approximately 9.8 m/s^2)
t is the time taken

Rearranging the formula to solve for time (t):

t^2 = (2 * h) / g

t = √((2 * h) / g)

Substituting the given values:
t = √((2 * 1.25 m) / 9.8 m/s^2)
t = √(0.2551)
t ≈ 0.505 s

So, it took approximately 0.505 seconds for the ball to reach the ground.

Now, to find the horizontal velocity at which the ball was thrown, we can use the formula for horizontal motion:

d = v_x * t

Where:
d is the horizontal distance (15 m in this case)
v_x is the horizontal velocity
t is the time taken (which we found to be 0.505 s)

Rearranging the formula to solve for horizontal velocity (v_x):

v_x = d / t

Substituting the given values:
v_x = 15 m / 0.505 s
v_x ≈ 29.70 m/s

So, the ball was thrown at approximately 29.70 m/s horizontally.

To find the vertical component of the ball's velocity when it hit the ground, we can use the formula for vertical motion:

v_y = g * t

Where:
v_y is the vertical velocity
g is the acceleration due to gravity (approximately 9.8 m/s^2)
t is the time taken (0.505 s)

Substituting the given values:
v_y = 9.8 m/s^2 * 0.505 s
v_y ≈ 4.95 m/s

So, the vertical component of the ball's velocity when it hit the ground is approximately 4.95 m/s.