Lab Report 2 (1).docx - Dennis To PHYS 1410L Sect 812...

• Lab Report
• 11
• 100% (1) 1 out of 1 people found this document helpful

This preview shows page 1 - 4 out of 11 pages.

Dennis To PHYS 1410L Sect. 812 Projectile Motion 6 March 2018 Jaden R., Kevin L., Andrew R. Objective: -The motion of the projectile in the gravitational field will be studied to gain an understanding of horizontal range, maximum height, time of flight, and trajectory of the projectile.
Introduction: At one point in our lives, we all must have seen balls being set into motion. We must have noticed that the motions of the ball form a curved path in the air. Balls that are set into motion mostly have a two-dimensional motion instead of one. In this experiment, the same type of motion will be analyzed. This motion is called a projectile motion, where the motion of the object is divided into two components(the x and y component). Two of the many important reasons why people study projectile motion are launching cannonballs at the right target during war and scoring goals in sports. A launched object has an initial velocity. The horizontal and vertical component of the velocity will be determined by using the following methods. V x = V cos θ V y = V sin θ Where: V x = horizontal component of velocity V = velocity V y = vertical component of velocity θ = angle that velocity makes with the horizontal Through studying the projectile motion, we will be able to discover many things. These include the time of flight of the object, the horizontal displacement, the vertical displacement, and the equation of the trajectory. As air resistance is neglected throughout the experiment, the object will only be accelerated by the acceleration due to gravity(g). The horizontal and vertical displacements can be determined by using the following Kinematic Equation. s = ut + 1 2 a t 2 Where: s = displacement u = initial velocity t = time a = acceleration For the horizontal motion : replace s with x, u with u cos θ The equation for the horizontal displacement will be x = u cos ( θ ) t + 1 2 a t 2 since air resistance is neglected and acceleration due to gravity only acts vertically, the acceleration for horizontal motion is 0. Therefore, the equation for horizontal displacement will be: x = u cos ( θ ) t (let this be Equation I)
For the vertical motion : replace s with y, u with u sin θ The equation for the vertical displacement will be y = u sin ( θ ) t + 1 2 a t 2 since air resistance is neglected, the acceleration of the object is only the acceleration due to gravity(g). Since y is measured upwards, the acceleration will be –g.Therefore, the equation
• • • 