A major leaguer hits a baseball so that it leaves the bat at a speed of 28.0 and at an angle of 36.9 above the horizontal. You can ignore air resistance.At what two times is the baseball at a height of 9.50 above the point at which it left the bat?
To find the two times when the baseball is at a height of 9.50 above the point at which it left the bat, we need to use the kinematic equations of motion.
First, let's break down the initial velocity of the baseball into its horizontal and vertical components. The horizontal component (Vx) remains constant throughout the motion, while the vertical component (Vy) is affected by gravity.
Given:
Initial speed (V) = 28.0 m/s
Launch angle (θ) = 36.9°
Height (y) = 9.50 m
Step 1: Resolve the initial velocity into its vertical and horizontal components:
Vx = V * cos(θ)
Vy = V * sin(θ)
Vx = 28.0 m/s * cos(36.9°) ≈ 22.41 m/s
Vy = 28.0 m/s * sin(36.9°) ≈ 16.92 m/s
Step 2: Determine the time it takes for the baseball to reach the highest point of its trajectory.
The vertical component of velocity at the highest point is zero, so we can use the following equation:
Vy = Vy0 - gt
where Vy0 is the initial vertical velocity and g is the acceleration due to gravity (-9.8 m/s²).
0 = Vy - gt
0 = 16.92 m/s - 9.8 m/s² * t
Solve for t:
t = 16.92 m/s / 9.8 m/s² ≈ 1.73 s
Step 3: Use the time found in step 2 to determine the time the baseball hits a height of 9.50 m above the point it left. We will use the equation:
y = y0 + Vy0t + 1/2 * g * t²
where y0 is the initial vertical position.
y = y0 + Vy0t + 1/2 * g * t²
9.50 m = 0 m + 16.92 m/s * t + 1/2 * (-9.8 m/s²) * t²
Rearrange the equation to form a quadratic equation:
4.9 m/s² * t² + 16.92 m/s * t - 9.50 m = 0
Solve the quadratic equation to find the time when the baseball is at a height of 9.50 m above the point it left. You can use the quadratic formula or factoring to solve for t.