A pitched ball is hit by a batter at a 45° angle and just clears the outfield fence, 90 m away. Assume that the fence is at the same height as the pitch and find the velocity of the ball when it left the bat. Neglect air resistance.
Well, if I were a pitcher and someone hit a ball that cleared the outfield fence, I would definitely be considering a career change. But anyway, let's do some math here.
First things first, we need to break down the motion of the ball into its horizontal and vertical components. Since the ball cleared the fence at a 45° angle, we know that the vertical and horizontal velocities are equal.
Now, assuming that the height of the fence is negligible compared to the distance to the fence, we can use basic kinematic equations to solve for the initial vertical velocity (Viy) and the time it takes for the ball to reach the fence (T).
Using the equation for vertical displacement:
90 m = Viy * T - (1/2) * g * T^2
Since we want to find the initial velocity, we can use the equation for vertical velocity:
Viy = g * T
Now, we can substitute the second equation into the first equation:
90 m = (g * T) * T - (1/2) * g * T^2
Simplifying the equation, we get:
90 m = (1/2) * g * T^2
Solving for T, we get:
T = sqrt(180 m / g)
Now that we have the time it takes for the ball to reach the fence, we can find the initial vertical velocity:
Viy = g * T
Substituting the value of T into the equation, we get:
Viy = g * sqrt(180 m / g)
Now, remember that the initial vertical velocity is equal to the initial horizontal velocity, so the velocity when the ball left the bat is:
V = sqrt((Vix)^2 + (Viy)^2)
Since Vix = Viy, we can simplify the equation to:
V = sqrt(2 * (g^2) * (180 m / g))
And finally, simplifying it further, we get:
V = sqrt(360 g m)
Now, I'm not really good at crunching numbers, but if you plug in the value of the acceleration due to gravity (g = 9.8 m/s^2), you should be able to find the velocity of the ball when it left the bat. Good luck, and don't forget to stretch before swinging that bat!
To find the velocity of the ball when it left the bat, we need to use the concepts of projectile motion.
Let's break down the given information:
- The distance from the bat to the fence is 90 m.
- The ball is hit at a 45° angle with respect to the ground.
- The fence is at the same height as the pitch.
Let's assume:
- The initial velocity of the ball when it left the bat is Vo.
- The horizontal component of the velocity (Vx) remains constant throughout the flight.
- The vertical component of the velocity (Vy) changes due to the acceleration due to gravity.
Now, using the kinematic equations, we can calculate the initial velocity of the ball.
1. Resolve the initial velocity (Vo) into horizontal (Vx) and vertical (Vy) components:
Vx = Vo * cos(45°) (the horizontal velocity remains constant)
Vy = Vo * sin(45°) (the vertical velocity changes due to gravity)
2. Use the equation for vertical displacement to find the time of flight (T):
-90 m = (1/2) * g * T² (where g is the acceleration due to gravity, approximately 9.8 m/s²)
T² = -180 / g
T = sqrt(-180 / g)
3. Use the equation for horizontal displacement to find Vx:
90 = Vx * T
Vx = 90 / T
4. Substitute the value of T into the equation for Vx to find Vy:
Vy = Vo * sin(45°)
Vo * sin(45°) = 90 / T
5. Substitute the value of T from step 2 into the equation for Vy to solve for Vo:
Vo * sin(45°) = (90 / sqrt(-180 / g))
Vo = (90 * sqrt(g) / sin(45°)) * sqrt(-180 / g)
6. Simplify the expression:
Vo = 90 * sqrt(2) * sqrt(g) / (sin(45°) * sqrt(180))
Vo = 90 * sqrt(2) * sqrt(9.8) / (sin(45°) * sqrt(180))
Vo ≈ 39.8 m/s
The velocity of the ball when it left the bat is approximately 39.8 m/s.
To find the velocity of the ball when it left the bat, we need to use the principles of projectile motion. Let's break down the problem into its components.
First, we need to decompose the initial velocity of the ball into its horizontal and vertical components. We know that the ball was hit at a 45° angle, so we can say that the initial velocity (v₀) can be split into two components: v₀x (horizontal component) and v₀y (vertical component).
Since there is no acceleration in the horizontal direction, v₀x remains constant throughout the motion. Therefore, v₀x = v₀ * cos(45°).
In the vertical direction, the ball experiences free fall motion, meaning there is a constant acceleration due to gravity acting downward. We can use the equation of motion to find v₀y:
v = v₀ + at
Since the ball just clears the fence, its final vertical position is zero. Therefore, we can use the equation y = v₀y * t - (1/2) * g * t², where y is the vertical displacement (height of the fence), g is the acceleration due to gravity, and t is the time of flight.
Since the ball was hit at a 45° angle, the time of flight (t) will be the same for both the upward and downward motion. Hence, we can also consider the time it takes for the ball to reach its maximum height (t/2) using the equation v = v₀ + at, where the final velocity (v) is zero at the highest point.
Using these equations, we can solve for the initial velocity (v₀).
Step 1: Find v₀x:
v₀x = v₀ * cos(45°)
Step 2: Find t/2:
At the highest point, v = v₀y - g * (t/2)
Since v = 0 at the highest point:
0 = v₀y - g * (t/2)
Step 3: Find t:
Rearrange the equation from Step 2 to solve for t:
t = 2 * v₀y / g
Step 4: Find v₀y:
Using the equation of motion, y = v₀y * t - (1/2) * g * t² and substituting t from Step 3, we get:
90m = v₀y * (2 * v₀y / g) - (1/2) * g * (2 * v₀y / g)²
Simplify this equation and solve for v₀y.
Step 5: Find v₀:
Using the value of v₀x from Step 1 and v₀y from Step 4, we can find v₀ by combining their components:
v₀ = √(v₀x² + v₀y²)
By following these steps and substituting the given values (angle = 45°, distance = 90m), you can calculate the velocity of the ball when it left the bat.
The range of projectile is
L=(v0)^2•sin2α/g.
If α=45 degr, sin2α = 1, then
L=v0^2/g.
v0=sqroot(L•g)= sqroot(90•9.8)= 29.7 m/s