Please Help!! A baseball clears a 3m tall wall positioned 120 m from home plate. The ball leaves the bat 1.2m above the ground and 45degrees above the horizontal. How fast was it going when it left the bat?

To determine the speed of the baseball when it left the bat, we can use the principles of projectile motion.

First, let's find the time it takes for the baseball to reach the wall. We can use the vertical component of the motion and the height of the wall to find the time taken.

The initial vertical velocity is given by V0y = V * sin(theta), where V is the initial velocity and theta is the angle of projection.

Using the equation for vertical displacement with constant acceleration:
y = V0y * t + 0.5 * a * t^2
where y is the height of the wall (3 m), V0y is the initial vertical velocity, a is acceleration due to gravity (-9.8 m/s^2), and t is the time taken.

Substituting the given values, we have:
3 = V * sin(45) * t + 0.5 * (-9.8) * t^2

Simplifying the equation, we get:
1.5t^2 - 1.41t - 3 = 0

Solving the quadratic equation, we find that t ≈ 2.49 seconds (rounded to two decimal places).

Next, we can use the horizontal component of the motion to find the initial velocity V.

The horizontal velocity V0x = V * cos(theta), where V is the initial velocity and theta is the angle of projection.

Using the equation for horizontal displacement with constant velocity:
x = V0x * t
where x is the distance from home plate to the wall (120 m), V0x is the initial horizontal velocity, and t is the time taken.

Substituting the known values and solving for V, we get:
120 = V * cos(45) * 2.49

Simplifying further, we have:
120 = V * 2.49 * 0.707
V ≈ 75.58 m/s (rounded to two decimal places).

Therefore, the speed of the baseball when it left the bat was approximately 75.58 m/s.