A pitched ball is hit by a batter at a 45-degree angle and just clears the outfield fence, 98 m away. If the fence is at the same height as the pitch, find the velocity of the ball when it left the bat. Ignore air resistance.
To find the velocity of the ball when it left the bat, we can use the range equation for projectile motion. The range equation is given by:
R = (v^2 * sin(2θ)) / g
Where:
R is the range (horizontal distance traveled),
v is the initial velocity of the projectile (ball),
θ is the angle at which the projectile is launched (45 degrees),
and g is the acceleration due to gravity (approximately 9.8 m/s^2).
In this case, we are given the range (98 m) and the launch angle (45 degrees). We need to solve the equation for the initial velocity (v).
Rearranging the equation, we have:
v^2 = (R * g) / sin(2θ)
Substituting the given values, we get:
v^2 = (98 * 9.8) / sin(90)
sin(90) is equal to 1, so the equation simplifies to:
v^2 = (98 * 9.8) / 1
v^2 = 960.4
To find the velocity, we take the square root of both sides:
v = √960.4
v ≈ 31 m/s
Therefore, the velocity of the ball when it left the bat was approximately 31 m/s.
To find the velocity of the ball when it left the bat, we can use the principles of projectile motion. In this case, we know the horizontal distance (range) the ball traveled (98 m) and the angle at which it was hit (45 degrees). We also know that the height of the fence is the same as the height of the pitch.
Here is how you can find the velocity of the ball when it left the bat:
Step 1: Split the initial velocity of the ball into its horizontal and vertical components. Let's assume that the initial velocity is V, and it can be split into Vx (horizontal component) and Vy (vertical component).
Step 2: Use trigonometry to find the value of the horizontal component (Vx). Since the ball was hit at a 45-degree angle, we know that the sine and cosine of 45 degrees are equal (sin 45 = cos 45 = 1/sqrt(2)). Therefore, Vx = V * cos 45.
Step 3: Determine the time it takes for the ball to reach the fence. Since we know the horizontal distance (range) traveled by the ball is 98 m and the horizontal component of velocity is Vx, we can use the equation: distance = velocity * time. Rearranging the equation, we get: time = distance / Vx.
Step 4: Use the time calculated in Step 3 to find the vertical component of velocity (Vy). Since the ball was hit at an angle, it will have some vertical velocity. The vertical distance covered by the ball can be calculated using the equation: distance = initial_velocity * time + (1/2) * g * time^2, where g is the acceleration due to gravity (approximately 9.8 m/s^2). However, since the height of the fence is the same as the height of the pitch, the vertical distance covered by the ball will be zero at the time it reaches the fence. So, distance = 0. Solving the equation gives us: 0 = Vy * time + (1/2) * g * time^2.
Step 5: Solve the equation from Step 4 to find the value of Vy. Since the term (1/2) * g * time^2 is zero, we can ignore it. Therefore, we get: 0 = Vy * time. This equation implies that either Vy is zero, which means the ball had no vertical velocity, or the time is zero, which means the ball never reached the fence. Since the ball just clears the fence, Vy cannot be zero, so we can conclude that the time is zero. Vy = 0.
Step 6: Calculate the magnitude of the initial velocity (V) using the horizontal (Vx) and vertical (Vy) components of velocity. V = sqrt(Vx^2 + Vy^2).
By following these steps, you should be able to find the velocity of the ball when it left the bat. Remember to substitute the known values into the equations as you go along.
how long is in the air?
hf=hi+Viv*time-1/2 g time^2
0=0+Vi*sin45*t-4.9 t^2
solve for t in terms of Vi
then horizontal equation
distance=vi*cos45*t put t in there, and solve for vi