A ball is thrown horizontally from a platform so that the initial height of the ball is 6.0 m above the level ground below. The ball lands 24 m from its original horizontal location. Find how fast the ball was thrown.

Show work i just don't understand

d = Vo*t + 0.5g*t^2 = 6 m.

0 + 4.9t^2 = 6,
t^2 = 1.22,
Tf = 1.11 s. = Fall time or time in flight.

Dx = Vo*Tf = 24 m,
Vo*1.11 = 24,
Vo = 21.7 m/s.

To find the initial velocity of the ball, you can use the equations of motion.

The ball is thrown horizontally, which means that the initial vertical velocity is zero. However, the ball is subject to the acceleration due to gravity (-9.8 m/s^2). The vertical displacement of the ball is given by the equation:

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

Since the initial vertical velocity is zero, the time it takes for the ball to reach the ground can be found by solving for t in the equation:

6 = (1/2) * 9.8 * t^2

Rearranging the equation gives:

2 * 6 = 9.8 * t^2
12 = 9.8 * t^2
t^2 = 12 / 9.8
t^2 ≈ 1.224

Taking the square root of both sides:

t ≈ √1.224
t ≈ 1.105 seconds

Now that we know the time it took for the ball to reach the ground, we can find the horizontal velocity using the equation:

v = d / t

where v is the velocity, d is the horizontal displacement, and t is time. Plugging in the given values:

v = 24 / 1.105
v ≈ 21.72 m/s

Therefore, the ball was thrown at a speed of approximately 21.72 m/s.

To find the initial speed at which the ball was thrown, we will use the equation of motion for horizontal projectile motion:

Range = Initial Velocity * Time

In this case, the range is given as 24 m, and the initial height is given as 6.0 m.

Since the ball is thrown horizontally, there is no vertical acceleration, and the only force acting on it is gravity. This means that the time taken for the ball to reach the ground will be the same as the time taken for the ball to travel horizontally.

To find the time, we can use the equation of motion for vertical projectile motion:

Vertical Displacement = Initial Velocity * Time + (1/2) * Acceleration * Time^2

In this case, the initial vertical displacement is 6.0 m and the acceleration is the acceleration due to gravity, which is approximately 9.8 m/s^2.

Simplifying the equation, we get:

6.0 m = 0 + (1/2) * 9.8 m/s^2 * Time^2

Rearranging the equation, we get:

Time^2 = (2 * 6.0 m) / 9.8 m/s^2
Time^2 = 1.224

Taking the square root of both sides, we get:
Time = 1.106 s

Now that we have the time, we can substitute this value back into the equation for horizontal motion:

24 m = Initial Velocity * 1.106 s

Rearranging and solving for the initial velocity, we get:

Initial Velocity = 24 m / 1.106 s

Calculating this, we find:
Initial Velocity = 21.68 m/s

Therefore, the ball was thrown with an initial speed of approximately 21.68 m/s.