Q1: What are the initial x & y components of the projectile's velocity? We can use trigonometry to solve for the initial x & y components of the projectiles velocity because we are given the angle 35 degrees above the horizontal and the hypotenuse of the triangle. To solve for the initial y component of the velocity, we use the sine function as follows: Sin 35° = y/(25m/s) y = (25m/s)(sin35° ) = 14.3m/s To solve for the initial x component of the velocity, we use the cosine function as follows: Cos 35° = x/(25m/s) x = (25m/s)(cos35° ) = 20.4m/s Q2: What are the x & y accelerations for the projectile? The acceleration in the y direction is equal to the gravitational field strength of the earth, which is 9.8m/s^{2}. The acceleration in the x direction is equal to 0m/s^{2} because we have assumed no air resistance. Q3: What is the maximum height of the projectile? Explain the theoretical assumption and the calculations you make to find this. In order to solve for the maximum height of the projectile (y_{max}), we must first solve for the time of flight of the projectile. Proceed to Q4 first and then solve for y_{max}. It turns out that the travel time of the projectile is 2.92 seconds. However, we are interested in the time when the cannonball is at the highest point in its trajectory. This occurs at t_{ 1/2}, or 1.46 seconds. We must employ the kinematics equation: x = v_{o}t + 1/2at^{2} y_{1}_{/2} = (v_{oy})(t_{1/2}) + 1/2(a_{y})(t^{2}) y_{ 1/2} = (14.3m/s)(1.46s) +1/2(9.8m/s^{2})(1.46s)^{2} y _{1/2} = 10.4 meters Q4: What is the total time of flight for this projectile? Explain the theoretical assumption and the calculation you make to find this. At the instant that the cannonball is at the top of its trajectory (when it has stopped rising, but before it begins its descent), its velocity in the y direction becomes zero. Because the trajectory of a parabola is symmetrical with respect to the yaxis, we know that this moment occurs halfway through the cannonball's flight (t _{1/2}). We can use this very important piece of information to solve for the time of flight of the cannonball. To do this we must employ the kinematics equation: v = v_{o} + at v _{1/2y} = v_{oy} + (a_{y})(t _{1/2}) (v _{1/2y}  v_{oy}) / (a_{y}) = t _{1/2} (0 m/s  14.3 m/s) / (9.8m/s^{2}) = t _{1/2 } t _{1/2} = 1.46 s Here, we have solved for the time it takes the cannonball to complete half of its flight. To solve for the total flight time, we must multiply this t _{1/2} by 2. total flight time = 2(t _{1/2}) = 2.92 seconds Q5: What is the total range traveled by this projectile? Explain the theoretical assumption and the calculation you make to find this. To solve for the range of the cannonball, we can employ the definition: distance = rate x time. To make this work out, we have to make the assumption that there is no air resistance. This ensures that the velocity in the x direction stays constant. We already solved for the initial velocity of the cannonball in the x direction and the total flight time. The sloution is as follows: distance = rate x time d = (20.4 m/s)(2.92 s) = 59.6 meters
