A baseball, hit 4 feet above the ground, leave the bat at an angle of 49 degrees and is caught by an outfielder 2.5 feet above the ground and 166 feet from home plate. Neglecting air resistance, what was the initial speed of the ball? Answer should be in feet per second.
+1982 We can solve this problem using the equations of motion for a projectile. Let's take the x-axis to be horizontal and the y-axis to be vertical.
The initial velocity of the ball can be decomposed into its x- and y-components:
v₀x = v₀ cos(49°)
v₀y = v₀ sin(49°)
where v₀ is the initial speed of the ball.
Using the fact that the ball was hit 4 feet above the ground, we can set the initial height y₀ to be 4 feet.
Using the fact that the ball was caught 2.5 feet above the ground, we can set the final height y equal to 2.5 feet.
Using the fact that the ball traveled 166 feet horizontally, we can set the horizontal displacement x equal to 166 feet.
Using the fact that there is no air resistance, we can set the acceleration due to gravity to be -32.2 feet per second squared.
Using the equations of motion for a projectile, we can write:
x = v₀x t
y = y₀ + v₀y t + (1/2)gt²
where t is the time of flight of the ball.
The time of flight can be found by setting y equal to 2.5 feet and solving for t:
2.5 = 4 + v₀ sin(49°) t + (1/2)(-32.2) t²
Solvingfor t, we get:
t = (v₀ sin(49°) + sqrt((v₀ sin(49°))² + 2(32.2)(4-2.5)))/32.2
Simplifying, we get:
t = (v₀ sin(49°) + sqrt((v₀ sin(49°))² + 25.84))/32.2
Using the fact that x = 166, we can write:
166 = v₀ cos(49°) t
Substituting the expression for t we derived earlier, we get:
166 = v₀ cos(49°) [(v₀ sin(49°) + sqrt((v₀ sin(49°))² + 25.84))/32.2]
Solving for v₀, we get:
v₀ = sqrt(166² (32.2)² / (cos²(49°) [(sin(49°))^2 + 25.84 / (32.2)^2]))
Simplifying, we get:
v₀ ≈ 125.4 feet per second
Therefore, the initial speed of the ball was approximately 125.4 feet per second.
