Projectile Motion Calculations PDF

## 3.3 Projectile Motion Calculations In Chapter 2, we used the kinematics equations for one-dimensional motions only. The key to solving two-dimensional problems is to break them up into two one-dimensional parts, then recombine them to produce a final answer. The problem set-up will now have a se...

## 3.3 Projectile Motion Calculations In Chapter 2, we used the kinematics equations for one-dimensional motions only. The key to solving two-dimensional problems is to break them up into two one-dimensional parts, then recombine them to produce a final answer. The problem set-up will now have a set of givens in the x direction and another set in the y direction. We will develop the process for solving such two-dimensional projectile problems in a series of steps. First, consider a horizontally fired cannon. We will set up the situation in Example 2, and then perform the calculations in Example 3. Next, consider a cannon that points upward at an angle in order to gain a greater range. Example 4 contains calculations to find the range. Example 5 calculates the maximum height achieved in the cannon shot. ### EXAMPLE 2 Setting up a projectile problem The Great Projecto, a circus duck, is shot at 40 m/s out of a horizontal cannon at the top of a cliff 100 m high. Derive the equations that represent the horizontal and vertical components of the motion. #### Solution and Connection to Theory We will assume a standard sign reference system. The first step is to set up givens for components in the x and y directions. * x) V1x = V2x = 40 m/s (horizontal velocity is constant) * ax = 0 * Adx = ? * At = ? * y) Viy = 0 * ay = -9.8 m/s² * Ady = -100 m. * At = ? Ady is negative because Projecto travels downward, the negative direction. The problem is now ready to be solved. The unknowns are time, and displacement in the horizontal direction, which is referred to as the range. Figure 3.15 shows Projecto shooting out with an initial horizontal velocity only as he sails through the air, he develops an ever-increasing vertical velocity downward. A diagram is shown with the following labels: 40 m/s, 100 m, + and 9.8 m/s². From the givens, you can see that there is a At in each dimension. Since this is the time from firing to landing, At is the same in both directions. So, if you can find a value for time in one of the directions, you can use that value in equations for the other direction. #### Choice of Equation The variables used in the x direction are ∆t, Ad, a, and v1. The equation containing all four of these variables is: △d = v1△t + 1/2at² This equation simplifies to Adx = vxAt because the acceleration in the x direction is zero and the velocity is constant. The variables in the y direction are the same as those in the x direction, hence we can use the same equation: Ady = viyat + ayť A diagram is shown with the following labels: METHOD of PROCESS, Given values, Split velocity into horizontal and vertical components, ax = 0, vx = constant, Adx = range, At, ay = -9.8 m/s², vy is changing, Ady = height, Adx = vxAt, Ady = vy • At + 1/2ayAt², Is At found in x?, NO, Calculate At using equation in y, Transfer to x direction, Find range, YES, Transfer to y direction, Find height Now let's finish the cannon problem. ### EXAMPLE 3 Finding the range The Great Projecto is shot out of a horizontal cannon at 40 m/s. If the cannon is sitting at the top of a cliff 100 m high, how far will the duck travel? #### Solution and Connection to Theory Given: * x) V1x = V2x = 40 m/s * ax = 0 * Adx = ? * At = ? * y) Viy = 0 * ay = -9.8 m/s² * Ady = -100 m * At = ? In the y direction, there is only one unknown. Solve in this direction first. Ady = v₁₁∆t + ay∆ť V₁₁ = 0, therefore At² = 2∆dy / ay At = ± √(2∆dy / ay) = ± √(2(-100 m) / -9.8 m/s²) = ±4.5 s Since time must be positive, At = +4.5 s. Now we use this value of time to calculate the range. Ad = vxAt ax = 0; therefore, Ad₁ = 40 m/s (4.5 s) = 180 m During the 4.5 s that Projecto took to fall 100 m in the y direction, he travelled 180 m in the x direction. ### EXAMPLE 4 Projection angled upward A cannonball is shot out of a cannon with a horizontal velocity component of 40 m/s and a vertical velocity component of 20 m/s [up]. If the cannon is sitting at the top of a cliff 100 m high, how far will the cannonball travel? #### Solution and Connection to Theory This problem is very much like Example 3, except it has a vertical velocity component. This means that the cannonball is being fired at an angle upwards from the cliff top. Given: * x) * V1x = V2x = 40 m/s * ax = 0 * Adx = ? * At = ? * y) * Viy = 20 m/s * ay = -9.8 m/s² * Ady = - 100 m * At = ? A diagram with the following labels is shown: VO, vx = 40 m/s, 100 m, -Range, + Notice in Fig. 3.17A that the height of the cliff is still 100 m, as it was in Examples 2 and 3. In this example, the object travelled higher than the cliff top, but the definition of displacement is final position minus initial position, which is just the height of the cliff. If we were to stop the problem at a different moment, then the displacement would be different. As before, solve first in the y direction. vy = 20 m/s #### QUADRATIC FORMULA Given the quadratic equation ax² + bx + c = 0, the solution is x = (-b ± √(b² - 4ac)) / (2a) #### ALTERNATIVE APPROACH Use v² = v1² + 2ay Ady to find v2y Then use V2y = v1y + ayat to find At. y) ∆dy = v1y∆t + 1/2ay∆ť -100 m = (20 m/s) ∆t + (-9.8 m/s²) Ať This is a quadratic equation in At. (-4.9 m/s²) At² + (20 m/s) ∆t + 100 m = 0 Using the quadratic formula, At = (-20 m/s ± √((20 m/s)² - 4(-4.9 m/s²) (100 m))) / (2(-4.9 m/s²)) At = (-20 m/s ± √(400 m²/s² + 1960 m²/s²)) / (-9.8 m/s²) At = (-20 m/s ± √(48.6 m/s²)) / -9.8 m/s² At = (-20 m/s + 48.6 m/s)/ -9.8 m/s² or (- 20 m/s - 48.6 m/s)/ -9.8 m/s² This gives At = 7.0 s and -2.9 s. Note that the units cancel to produce the correct unit for At. A negative time does not apply in real life, so the correct value for At is 7.0 s. Now substitute 7.0 s for At in the equation for the x direction. ax = 0; therefore, Adx = v1x At Ad = 40 m/s (7.0 s) = 280 m Because this projectile spent more time in the air than the one in Example 3, it managed to travel a greater distance, 280 m instead of 180 m. #### NEGATIVE TIME We can give a meaning to the negative time by illustrating the problem using a parabola. The quadratic equation for At describes a motion for a vertical displacement of 100 m for the initial velocity components given in the problem. Those same values could also be achieved by starting at the ground 2.9 s earlier, and firing a projectile upward. After 2.9 s, the upward velocity of the projectile would be 20 m/s. Mathematically, that earlier portion of the motion is the part of the parabola to the left of the vertical axis, as shown in Fig. 3.17B. A diagram with the following labels is shown: 150 m Δαd, 100 m, 50 m, -4 s, -2 s, 0, 2 s, 4 s, 6 s, At, 8 s To find the maximum height of the shot, modify the solution as shown in Example 5. ### EXAMPLE 5 Maximum height of a projectile For the cannonball in Example 4, find its maximum height. #### Solution and Connection to Theory Given: We need the y direction only because the maximum height involves the displacement in the y direction only. y) * Viy = 20 m/s * V2y = 0 m/s * ay = -9.8 m/s² * Ady = ? * At = ? The final velocity is zero because at the maximum height, the cannonball stops moving up. The formula of choice for this case is v²2, = v²1, + 2aΔd Then Ady = (v²2, - v²1,) / 2a = (0 m/s)² – (20 m/s)² / 2(-9.8 m/s²) = -400 m²/s² / -19.6 m/s² = 20 m The cannonball will rise 20 m above its original location. A diagram is shown with the following labels: METHOD of PROCESS, Given values, y ) Ady = v1yAt + 1/2ayAt², Is maximum height required?, NO, Solve projectile problem as in Fig. 3.16, YES, Set v2y = 0, Use kinematics equations, Find At, Find dy ### Finding Final Velocity To find the final velocity of the projectile as it hits the ground, we need to find the values of the two vector components that comprise it. In the x direction, the final velocity is the same as the initial velocity because there is no horizontal acceleration. Then we calculate the final velocity in the y direction from the given data. We get the resultant final velocity by adding the two components using the head-to-tail method. With a scale diagram, you can measure the magnitude and direction of the final velocity. For a more precise value, use Pythagoras’ theorem and the inverse tangent of the ratio of the magnitudes of the components. ### EXAMPLE 6 Finding the final velocity What is the final velocity of the Great Projecto, if he is shot out of a cannon with a horizontal velocity component of 19 m/s and a vertical component of 23 m/s? For his stunt, he lands in a net 2.0 m above the point of launch and 70 m away (Fig 3.20A). A diagram with the following labels is shown: Fig 3.20A, V1y, Vx, V2y, Ady, Range, Fig 3.20B, V2y, Vx, 0, VT, V2x For image, see student text. #### Solution and Connection to Theory Given: * x) vx = 19 m/s. Because the velocity is constant in the x direction, this value is also the final x velocity. * y) ay = -9.8 m/s² * Ady = 2.0 m * V1 = 23 m/s * V = ? The formula of choice is V2 = √(V²1 + 2aΔd) = ? V2 = ±√(23 m/s² + 2 (-9.8 m/s²) 2.0 m) = ± 22 m/s We obtain two answers (a positive and a negative) because Projecto is two metres above the point of launch at two different times. Just after leaving the cannon (up two metres vertically), his vertical velocity is 22 m/s [up]. As he reaches the net, he has a vertical velocity of 22 m/s [down]. Here, we are interested in the final velocity only. Because Projecto is moving downward, the velocity is negative. Therefore, V2y = -22 m/s. Now we combine the x and y components of the final velocity from Fig 3.20A by connecting them head to tail. For image, see student text. In Fig 3.20B, the final velocity vector was obtained by adding the two components. The resulting vector diagram is a right-angle triangle. Using Pythagoras' theorem, VT = √(19 m/s)² + (-22 m/s)² = 29 m/s If you used a scale diagram, then measure the angle. If not, use the tangent function, substituting the magnitudes of the components. 0 = tan−1 (22 m/s / 19 m/s) = 49° To find the compass directions, we check the signs of the components, vx (+) and v2y (-). The signs tell us that Projecto was moving to the right and down. His final velocity is 29 m/s [R49°D]. A diagram is shown with the following labels: Fig 3.21 Finding Final Projectile Velocity, x) V1x = constant, V2x = V1x, y) ay, v1y, Ady, V2y =Vv1y² + 2ayAdy, V2x =VV2x² + v2y , 0 = tan-1 (V2y / V2x) , Fig 3.22 Complete Projectile Overview, Givens, x, y, ax = 0, Adx, V1x = √2x, y, ay = -9.8 m/s², Ady, Vly, Ad = v₁ At + 1/2 at², Can you calculate time in x ?, NO, Calculate time in y, YES, Complete the initial problem, Calculate V2y, Vf = √v2x² + v2y , 0 = tan-1(V2y / V2x), PROCESS of THOUGHT, PUTTING it all together