A baseball was hit by the batter at an angle of 30o. The baseball reached a distance of 15 m, and a maximum height of 6 meters. Find the initial velocity of the ball.
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A baseball was hit by the batter at an angle of 30o. The baseball reached a distance of 15 m, and a maximum
Answer:
18.04 m/s
Explanation:
To solve this problem, we need to use the kinematic equations of motion for projectiles. We know the initial angle (θ), the initial velocity (v₀), the maximum height (yᴍ), and the horizontal distance traveled (x). We want to find the initial velocity of the ball.
First, let's find the time of flight (t) of the ball using the equation for maximum height:
yᴍ = (v₀² sin²θ)/(2g)
where g is the acceleration due to gravity, which is approximately 9.81 m/s².
Plugging in the given values, we get:
6 m = (v₀² sin²30°)/(2 × 9.81 m/s²)
Simplifying and solving for v₀, we get:
v₀ = √((2gyᴍ)/(sin²θ)) = √((2 × 9.81 m/s² × 6 m)/(sin²30°)) = 19.62 m/s
Now, let's find the time of flight (t) of the ball using the equation for horizontal distance:
x = v₀ cosθ × t
Plugging in the given values, we get:
15 m = v₀ cos30° × t
Simplifying and solving for t, we get:
t = 15 m/(v₀ cos30°) = 15 m/(19.62 m/s × cos30°) = 1.53 s
Finally, we can use the time of flight to find the initial velocity (v₀) using the equation for vertical motion:
y = v₀ sinθ × t - (1/2)gt²
Plugging in the given values and solving for v₀, we get:
v₀ = yᴍ/(sinθ × t) + (1/2)gt = 6 m/(sin30° × 1.53 s) + (1/2) × 9.81 m/s² × 1.53 s = 18.04 m/s
Therefore, the initial velocity of the ball was approximately 18.04 m/s.