Introduction to Mechanics Range of a Projectile Trajectory Equation More Examples Lana Sheridan De Anza College Oct 25, 2017
Last time max height of a projectile time-of-flight of a projectile range of a projectile
Overview maximizing the range of a projectile investigating projectile motion more examples
Range of a Projectile ely, with time t as the coms in its parabolic path e velocity and accelerather? (a) nowhere (b) the choices, at what point are parallel to each other? y S vi v y 0 Projectile special case of projectile ed from the origin at t i 5 nd returns to the same horiseballs, footballs, and golf aunched. analyze: the peak point, t, which has coordinates projectile, and the distance ly in terms of v i, u i, and g. O ui h R x Figure 4.9 A projectile launched over a flat x surface = vfrom x t the origin at t i 5 0 with an initial ( velocity ) S vi. The maximum height 2v0 sin of the θ R projectile = v i is cos h, and θ the horizontal range is R. At, the peak gof the trajectory, the particle has coordinates = 2v (R/2, R 0 2 sin θ cos θ h). g R = v 0 2 sin(2θ) g
Maximizing Range R = v 2 0 sin(2θ) g What angle maximizes the range of the projectile?
Maximizing Range ATICS R = v 2 0 sin(2θ) g What angle maximizes the range of the projectile? Height, y (m) 15 12.5 10 7.5 5 = 60 = 30 = 45 2.5 degrees) 60 75 90 O 10 20 Distance, x (m) (b) e in the absence R is maximized of air resistance when sin(2θ) = 1 θ = 45 or a projectile launched with an initial speed of 20 m/s. Note that the maximum range y greater than or less than 45, such as 30 and 60, give the same range. (b) Trajectories of 30 40
Projectile Trajectory Suppose we want to know the height of a projectile (relative to its launch point) in terms of its x coordinate. Suppose it is launched at an angle θ above the horizontal, with initial velocity v i.
Projectile Trajectory Suppose we want to know the height of a projectile (relative to its launch point) in terms of its x coordinate. Suppose it is launched at an angle θ above the horizontal, with initial velocity v i. For the x-direction: x = v i cos θt t = x v i cos θ
Projectile Trajectory Suppose we want to know the height of a projectile (relative to its launch point) in terms of its x coordinate. Suppose it is launched at an angle θ above the horizontal, with initial velocity v i. For the x-direction: y-direction: Substituting for t gives: x = v i cos θt t = y = v i sin θt 1 2 gt2 x v i cos θ g y = (tan θ)x 2vi 2 cos 2 θ x 2
Projectile Trajectory ( y = (tan θ)x g 2vi 2 cos 2 θ ) x 2 This is why projectiles trace out a parabola with respect to horizontal position, as well as with respect to time.
Shepard sim Golf on the Moon with him o The projectile s path without air resistance is a symmetricalfra Mauro parabola. His distanc sand trap. With air resistance, this is no longer the case. Now, w that R varie y 3.5 = 45 when sin 2u range. Thus 3 2.5 = 35 What Happens with Air Resistance? 2 1.5 1 0.5 O = 25 = 15 x 2 4 6 8 10 Range (m) FIGURE 4 9 Projectiles with air 1 Figure fromresistance Walker, Physics, page 97. As expecte depend str 2 tional to v 0 Note th at the same
Validity of Range Equation Expression for the range of a projectile: When is it valid? R = v 2 0 sin(2θ) g
Validity of Range Equation Expression for the range of a projectile: R = v 2 0 sin(2θ) g When is it valid? when the projectile lands at the same height it is launched from! when there is no air resistance (we will not deal with air resistance in this course)
eight h does b Figure P4.24 Projectile Motion Example a 25. A playground is on the flat roof of a city school, 6.00 m above the street below (Fig. P4.25). The vertical wall of the building is h 5 7.00 m high, forming a 1-m-high railing around the playground. A ball has fallen to the street below, and a passerby returns it by launching it at an angle of u 5 53.0 above the horizontal at a point d 5 24.0 m from the base of the building wall. The ball takes 2.20 s to reach a point vertically above the wall. (a) Find the speed at which the ball was launched. (b) Find the vertical distance by which the ball clears the wall. (c) Find the horizontal distance from the wall to the point on the roof where the ball lands. P4.22 h u d Figure P4.25 1 Serway & Jewett, Physics for Scientists & Engineers, 9th ed, #25, p103.
Projectile Motion Example Part (a) Hypothesis: about 12 m/s (faster than a normal person can run) Given: x = 24.0 m, t = 2.20 s, θ = 53.0 Asked for: v 0
Projectile Motion Example Part (a) Hypothesis: about 12 m/s (faster than a normal person can run) Given: x = 24.0 m, t = 2.20 s, θ = 53.0 Asked for: v 0 Strategy: we know x = v 0x t and v 0x = v 0 cos θ. Rearranging: x = v 0 cos θt v 0 = x cos θt = (24.0 m) cos(53 )(2.20 s) = 18.1 m/s
Projectile Motion Example Part (b) Hypothesis: about 2 m Given: x = 24.0 m, t = 2.20 s, θ = 53.0, v 0 = m/s, h = 7.00 m Asked for: height above the wall, y h
Projectile Motion Example Part (b) Hypothesis: about 2 m Given: x = 24.0 m, t = 2.20 s, θ = 53.0, v 0 = m/s, h = 7.00 m Asked for: height above the wall, y h Strategy: there are a couple of ways to solve it. One way: y = v 0y t 1 2 gt2 y h = v 0 sin θt 1 2 gt2 h = (18.1 m/s) sin(53 )(2.20 s) 1 2 (9.81)(2.20)2 7m = 1.11 m
Projectile Motion Example Part (c) Hypothesis: again, about 2 m Given: θ = 53.0, v 0 = m/s, y = 6.00 m Asked for: distance behind the wall, x d
Projectile Motion Example Part (c) Hypothesis: again, about 2 m Given: θ = 53.0, v 0 = m/s, y = 6.00 m Asked for: distance behind the wall, x d Strategy: Could find t, then x. Or, use trajectory equation: g y = (tan θ)x 2vi 2 cos 2 θ x 2
Projectile Motion Example Solve for x. g 2vi 2 cos 2 θ x 2 (tan θ)x + y = 0
Projectile Motion Example Solve for x. g 2vi 2 cos 2 θ x 2 (tan θ)x + y = 0 tan θ ± x = ( tan 2 θ 4 ( ) g 2 2v 2 i cos 2 θ g 2vi 2 cos 2 θ ) y
Projectile Motion Example Solve for x. g 2vi 2 cos 2 θ x 2 (tan θ)x + y = 0 tan θ ± x = ( tan 2 θ 4 ( ) g 2 2v 2 i cos 2 θ Putting in the numbers in the question: g 2vi 2 cos 2 θ x = 26.79 m or x = 5.44 m ) y Want the solution that is larger than 24 m, since ball makes it onto roof. Distance behind wall: 26.79 m - 24 m = 2.79 m.
Relative Motion And Projectile Motion /s releases a ball from a height of 1.25 m above the ground. Given = 0.250 Observer s and (b) on the t = skateboard 0.500 s. (c) sees Find the the ball velocity, fall straight speed, down. and direc- ve- and due ves y v 0 g h = 1.25 m gt 2, ons. -gt. O continued on next page Another observer on the sidewalk sees the ball as a horizontally launched projectile. x
Relative Motion And Projectile Motion #73, page 108 To decide who pays for lunch, a passenger on a moving train tosses a coin straight upward with an initial speed of 4.38 m/s and catches it again when it returns to its initial level. From the point of view of the passenger, then, the coin s initial velocity is (4.38 m/s) ŷ. The train s velocity relative to the ground is (12.1 m/s) ˆx. (a) What is the minimum speed of the coin relative to the ground during its flight? At what point in the coin s flight does this minimum speed occur? Explain.
Relative Motion And Projectile Motion #73, page 108 To decide who pays for lunch, a passenger on a moving train tosses a coin straight upward with an initial speed of 4.38 m/s and catches it again when it returns to its initial level. From the point of view of the passenger, then, the coin s initial velocity is (4.38 m/s) ŷ. The train s velocity relative to the ground is (12.1 m/s) ˆx. (a) What is the minimum speed of the coin relative to the ground during its flight? At what point in the coin s flight does this minimum speed occur? Explain. 12m/s, At the top of its path, where the y-component of velocity is zero.
Relative Motion And Projectile Motion #73, page 108 To decide who pays for lunch, a passenger on a moving train tosses a coin straight upward with an initial speed of 4.38 m/s and catches it again when it returns to its initial level. From the point of view of the passenger, then, the coin s initial velocity is (4.38 m/s) ŷ. The train s velocity relative to the ground is (12.1 m/s) ˆx. (b) Find the initial speed and direction of the coin as seen by an observer on the ground. v 0 = v 2 0x + v 2 0,y ( ) θ = tan 1 v0y v 0x v 0 = 12.9 m/s, at 19.9 above the horizontal
Relative Motion And Projectile Motion #73, page 108 To decide who pays for lunch, a passenger on a moving train tosses a coin straight upward with an initial speed of 4.38 m/s and catches it again when it returns to its initial level. From the point of view of the passenger, then, the coin s initial velocity is (4.38 m/s) ŷ. The train s velocity relative to the ground is (12.1 m/s) ˆx. (b) Find the initial speed and direction of the coin as seen by an observer on the ground. v 0 = v 2 0x + v 2 0,y ( ) θ = tan 1 v0y v 0x v 0 = 12.9 m/s, at 19.9 above the horizontal
Relative Motion And Projectile Motion #73, page 108 To decide who pays for lunch, a passenger on a moving train tosses a coin straight upward with an initial speed of 4.38 m/s and catches it again when it returns to its initial level. From the point of view of the passenger, then, the coin s initial velocity is (4.38 m/s) ŷ. The train s velocity relative to the ground is (12.1 m/s) ˆx. (b) Find the initial speed and direction of the coin as seen by an observer on the ground. v 0 = v 2 0x + v 2 0,y ( ) θ = tan 1 v0y v 0x v 0 = 12.9 m/s, at 19.9 above the horizontal
Relative Motion And Projectile Motion #73, page 108 To decide who pays for lunch, a passenger on a moving train tosses a coin straight upward with an initial speed of 4.38 m/s and catches it again when it returns to its initial level. From the point of view of the passenger, then, the coin s initial velocity is (4.38 m/s) ŷ. The train s velocity relative to the ground is (12.1 m/s) ˆx. (c) Use the expression for h = y max to calculate the maximum height of the coin, as seen by an observer on the ground. h = 0.978 m h = v 2 0y 2g
Relative Motion And Projectile Motion #73, page 108 To decide who pays for lunch, a passenger on a moving train tosses a coin straight upward with an initial speed of 4.38 m/s and catches it again when it returns to its initial level. From the point of view of the passenger, then, the coin s initial velocity is (4.38 m/s) ŷ. The train s velocity relative to the ground is (12.1 m/s) ˆx. (c) Use the expression for h = y max to calculate the maximum height of the coin, as seen by an observer on the ground. h = 0.978 m h = v 2 0y 2g
Relative Motion And Projectile Motion #73, page 108 To decide who pays for lunch, a passenger on a moving train tosses a coin straight upward with an initial speed of 4.38 m/s and catches it again when it returns to its initial level. From the point of view of the passenger, then, the coin s initial velocity is (4.38 m/s) ŷ. The train s velocity relative to the ground is (12.1 m/s) ˆx. (c) Use the expression for h = y max to calculate the maximum height of the coin, as seen by an observer on the ground. h = 0.978 m h = v 2 0y 2g
Relative Motion And Projectile Motion #73, page 108 To decide who pays for lunch, a passenger on a moving train tosses a coin straight upward with an initial speed of 4.38 m/s and catches it again when it returns to its initial level. From the point of view of the passenger, then, the coin s initial velocity is (4.38 m/s) ŷ. The train s velocity relative to the ground is (12.1 m/s) ˆx. (d) Calculate the maximum height of the coin from the point of view of the passenger, who sees only one-dimensional motion. h = 0.978 m
Relative Motion And Projectile Motion #73, page 108 To decide who pays for lunch, a passenger on a moving train tosses a coin straight upward with an initial speed of 4.38 m/s and catches it again when it returns to its initial level. From the point of view of the passenger, then, the coin s initial velocity is (4.38 m/s) ŷ. The train s velocity relative to the ground is (12.1 m/s) ˆx. (d) Calculate the maximum height of the coin from the point of view of the passenger, who sees only one-dimensional motion. h = 0.978 m
Summary range of a projectile trajectory equation for a projectile Homework Walker Physics: PREV: Ch 4, onward from page 100. Problems: 37, 47, 49, 71