A baseball player hits a home run that just barely clears the 12m high fence in right field, 94m from home plate. If the ball was at a height of 1m when hit, and its initial path after it left the bat was at an angle of 35 degrees above the horizontal, what was the ball's initial speed when it left the bat?

Question

A baseball player hits a home run that just barely clears the 12m high fence in right field, 94m from home plate. If the ball was at a height of 1m when hit, and its initial path after it left the bat was at an angle of 35 degrees above the horizontal, what was the ball's initial speed when it left the bat?

Answer 1

break vi into horizontal and vertical components

94=v*cos35*t in the horizontal, find t, the time in air.
Vertical
12=1+vsin35*t-4.9t^2 solve for v, knowing t. Use the quadratic equation.

Answer 2

You will likely not need to use the quadratic equation.

Answer 3

To find the initial speed of the ball when it left the bat, we can use the principle of projectile motion and break down the problem into horizontal and vertical components.

First, let's analyze the vertical motion of the ball. The ball's initial height is 1m, and it needs to clear a 12m high fence. This means the ball needs to reach a maximum height of 12m + 1m = 13m.

The maximum height of a projectile occurs when the vertical velocity component becomes zero. In this case, the final vertical velocity is zero when the ball reaches its maximum height. We can use the equation:

Vfy = Viy + at

Where:
Vfy = final vertical velocity (0 m/s in this case)
Viy = initial vertical velocity
a = acceleration due to gravity (-9.8 m/s², assuming negligible air resistance)
t = time taken to reach the maximum height

Since the ball is launched at an angle of 35 degrees above the horizontal, we can find the initial vertical velocity using trigonometry:

Viy = Vi * sin(θ)
where θ is the launch angle (35 degrees) and Vi is the initial velocity.

Now, let's analyze the horizontal motion of the ball. The horizontal velocity remains constant throughout the motion (assuming no air resistance). We can use the equation:

Vfx = Vix

where:
Vfx = final horizontal velocity
Vix = initial horizontal velocity

Since there is no acceleration in the horizontal direction, we can use the equation:

Vix = Vi * cos(θ)

Now, we have two equations:

Viy = Vi * sin(θ)
Vix = Vi * cos(θ)

We can solve for Vi by substituting these values into the equation for the vertical motion:

0 = Vi * sin(θ) + (-9.8 m/s²) * t

Since the ball reaches the maximum height when the vertical velocity is zero, we can determine the time taken to reach the maximum height:

0 = Vi * sin(θ) + (-9.8 m/s²) * t
t = Vi * sin(θ) / 9.8 m/s²

We can substitute this value of t back into the equation for horizontal motion:

D = Vix * t
where D is the horizontal distance (94m in this case)

94m = (Vi * cos(θ)) * (Vi * sin(θ) / 9.8 m/s²)

Now we have an equation with only a single unknown, Vi. We can rearrange and solve for Vi.

Vi = sqrt((94m * 9.8 m/s²) / (sin(2θ)))

Plugging in the values given:
θ = 35 degrees

Vi = sqrt((94m * 9.8 m/s²) / (sin(2 * 35 degrees)))

Solving this equation will give us the initial speed of the ball when it left the bat.