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Unit Four: Projectile/Falling-Body Motion 1 The first three formulas concern vertical motion; the fourth concerns horizontal motion. Formula 1: the distance a body falls from rest because of gravity y = ½ at2 Formula 2: the distance a body falls from gravity with initial vertical velocity v0: y = ½ at2 + v0t Unit Four : Projectile/Falling-Body Motion 2 Formula 3: the height of a body falling from gravity and initial vertical velocity starting at an initial height, y0: y = ½ at2 + v0t + y0 (y and y0 are in feet; t is in seconds; v0 is in feet/second; a is −32 feet/sec.2; downward motion is negative) Formula 4: (horizontal motion) the horizontal distance traveled by a body moving with an initial horizontal velocity (vH0) is as follows: x = vH0t (x is in feet; t is in seconds, and vH0 is in feet/second) Unit Four: Projectile Motion – Problem 1 Problem 1: Standing near the edge of the roof of a building, Frank throws a baseball from a height of 100 feet upwards with a vertical velocity of 25 feet per second a height of 100 feet at a velocity. (As it falls, the ball will fall past the roof down toward the ground.) How high above the ground is the baseball after three seconds? Use the seven steps to solve this problem. If you cannot, look at the solution shown below. This is not really Frank . . . . . . .but it’s who he’s pretending to be. (I know, Frank is throwing with a vertical velocity, but he’s pretty confused; don’t give him a hard time.) Unit Four: Projectile Problem 1- Solution I • Read the problem. • Organize the data: § The initial height of the baseball is 100 feet above the ground. § The vertical velocity is 25 feet/second. § The time elapsed is three seconds . • The problem asks for the height of the baseball at that time (after 3 seconds)? That variable is best expressed as y, which the projectile formulas use as the vertical height. Unit Four: Projectile Problem 1- Solution II 4. We have three formulas from which to choose for vertical height. This problem involves not only the effect of gravity, but also an initial velocity and an initial height, given all of which we are to find the height at a given time. Only Formula 3 seems to fit this problem, so we choose it: y = ½ at2 + v0t + y0 Unit Four: Projectile Problem 1- Solution III 5. Substitute the given data into the formula we have chosen: y = ½ (−32)32 + 25(2) + 100 6. Solve mathematically: y = ½(− 32)(9) + 50 + 100 y = − 144 + 150 y=6 Solution: after 3 seconds, the ball is at a height of 6 feet. 7. If you check the units of measurement in each term in the formula, you will see that each separate term is in feet. In the first term on the right of the equation, for example, feet/sec.2 is multiplied by sec.2, so that the unit of measurement for the entire term is feet. Unit Four: Projectile Problem 2 Problem 2: Robin stands near the edge of a cliff in the desert. She shoots an arrow into the air from a height of 100 feet above the ground below with a vertical velocity of 40 feet per second a height and a horizontal velocity of 30 feet per second (sending it over the cliff). What will be the height of the arrow relative to the ground below the cliff when it has traveled horizontally 75 feet? Robin Practices! “I shot an arrow into the air. . . . . . Unit Four: Projectile Problem 2 – Solution I • We first read the problem until we understand what tells and asks us. • We organize the problem data: § The object (arrow) begins at a height of 100 feet relative to the ground below, § a vertical velocity of 40 feet per second, § a horizontal velocity of 30 feet per second, § and travels horizontally a total of 75 feet. Unit Four: Projectile Problem 2 – Solution II 3. We are asked what the height of the arrow is when it has traveled 75 feet horizontally. § We are missing the amounts of two quantities: the length of time the arrow travels to go 75 feet horizontally AND the height of the arrow at that time. § We designate t as the length of time and y as the height of the arrow (following the conventions in the Formulas). 4. We are asked to find the height, y, of the arrow. Math Formula 3 seems most appropriate to find y because we have both an initial height and an initial vertical velocity. However, we cannot immediately use Formula 3 because we do not yet know t, which depends on the horizontal distance. To calculate t first, we need to use Math Formula 4. Unit Four: Projectile Problem 2 – Solution III 5. & 6. Substep A. We substitute the data we are given into Formula 4 and solve for t: x = vH0t 75 feet = (30 feet/second)t t = 2.5 seconds Substep B. We take Formula 3, substitute the given data and t = 2.5, and solve for y: y = ½ at2 + v0t + y0 Unit Four: Projectile Problem 2 – Solution IV 5. & 6. (cont.). y = ½ at2 + v0t + y0 y = ½(−32ft/sec.2)(2.5sec.)2 + 40ft./sec.(2.5sec.) + 100 feet y = (−16ft/sec.2)(6.25sec2) + 100 feet + 100 feet y = −100 feet + 100 feet + 100 feet y = 100 feet The height is 100 feet above ground at that point. 7. We check the units which we calculated as we proceeded, and they are correct. Other Projectile/Falling-Body Problems ØOther projectile/falling-body problems are supplied in the accompanying written materials for you to try. If you have difficulty, guidance is available from tutors or in written form.
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