An 11-kg bowling ball is thrown up a ramp, as shown. If we have a coefficient of dynamic friction of 0.11, determine: a) How high does the bowling ball go? Can you fall over the edge of the ramp? b) At what time does it reach its maximum height? Other data: |v|=43 m/s x=31m angle=11 degrees

To solve this problem, we can break it down into two parts: finding the maximum height the bowling ball reaches and determining the time it takes to reach that height. We'll use the given data and apply principles of physics. Let's start with part a: a) How high does the bowling ball go? Can it fall over the edge of the ramp? First, we need to find the initial vertical velocity of the bowling ball. We can use trigonometry to find the vertical component of the initial velocity: Given: The magnitude of the velocity (v) is 43 m/s. The angle (θ) of the ramp is 11 degrees. Vertical velocity (Vy) = v * sin(θ) Vy = 43 m/s * sin(11 degrees) Vy ≈ 43 m/s * 0.1908 ≈ 8.2 m/s Now, let's calculate the initial kinetic energy (K.E) of the ball: K.E = 0.5 * m * v^2 K.E = 0.5 * 11 kg * (43 m/s)^2 K.E ≈ 0.5 * 11 kg * 1849 m^2/s^2 ≈ 10169.5 J Now, let's calculate the work done against friction as the ball moves up the ramp: Work = Frictional Force * Distance Work = μ * m * g * d Where: μ (mu) is the coefficient of dynamic friction (given as 0.11). m is the mass of the ball (11 kg). g is the acceleration due to gravity (approximately 9.81 m/s^2). d is the distance along the ramp (given as 31 m). Work = 0.11 * 11 kg * 9.81 m/s^2 * 31 m Work ≈ 367.9731 J Now, let's find the initial potential energy (P.E) of the ball: P.E = m * g * h Where: h is the height we want to find. We can now set up an energy conservation equation: Initial K.E - Work = Initial P.E 10169.5 J - 367.9731 J = 11 kg * 9.81 m/s^2 * h Solving for h: 10132.64 J = 107.91 m/s^2 * h h ≈ 10132.64 J / (107.91 m/s^2) h ≈ 90.83 meters The bowling ball reaches a height of approximately 93.68 meters. Now, let's check if it falls over the edge of the ramp: The vertical height of the ramp (h_ramp) can be calculated using trigonometry: h_ramp = d * sin(θ) h_ramp = 31 m * sin(11 degrees) h_ramp ≈ 31 m * 0.1908 ≈ 5.92 meters Since the ball reached a height of approximately 93.68 meters, which is much higher than the height of the ramp (5.92 meters), it will not fall over the edge of the ramp. Now, let's move on to part b: b) At what time does it reach its maximum height? To find the time it takes to reach maximum height, we can use the following kinematic equation: Vy = V0y + at Where: Vy is the final vertical velocity at maximum height (0 m/s since it momentarily stops). V0y is the initial vertical velocity (8.2 m/s). a is the acceleration due to gravity (-9.81 m/s^2, taking it as negative because it's acting downward). t is the time we want to find. 0 m/s = 8.2 m/s - 9.81 m/s^2 * t Solving for t: 9.81t = 8.2 t ≈ 8.2 m/s / 9.81 m/s^2 ≈ 0.836 seconds So, it takes approximately 0.836 seconds for the bowling ball to reach its maximum height.