Review | Constants Part A An archer standing on a 15° slope shoots an arrow 20 above the horizontal, as shown in the figure (Figure 1) How far down the slope does the arrow hit if it is shot with a speed of 50 m/s? Hint: There are several strategies to solve this problem. This hint discusses two different methods. Figure Method 1: 1 of 1 The arrow must land on the slope, so write y(x) of the slope as y= tan(phi)*x, where phi is angle of the slope and it must be negative. Note that you need to define a coordinate system, which you might assign the origin at the position of the archer, and +x is to the right in graphic. All angles must be with respect to the positive x direction. The slope angle in the picture does not conform to this requirement so you have to determine the slope angle from the information given. That is why you should write -phi for the angle of the slope 200 The distance down the slope is just the sqrt[(displacement in x)^2+(displacement in y)^2]. Strategy: write the usual kinematic equations for x(t) and y(t) with v0 and launch 15° onclo thato ninnn hath aro givon to vou C.htit+n ult) uith tho acuotion tonlohi\*, +n ont y)^2]. Strategy: write the usual kinematic equations for x(t) and y(t) with v0 and launch angle theta since both are given to you. Subtitute y(t) with the equation tan(phi)*x to get two equations for x as a function of t. From those two equations, solve for time of flight, t of the arrow. Once t is known, you can solve for displacement in x from the usual kinematic equation x(t) and you obtain the displacement in y from the usual kinematic equation for the y-component. Method 2: If you use a rotated coordinate system, so that +x lies in the down direction of the slope, then x is the distance. However in this coordinate system, gravitational acceleration has both x and y components, so this method requires that you solve the kinematic equations for acceleration in both y and x. Determine and x and y components for gravitational acceleration, call them g_x and g_y. You should be able to use geometry to determine that g_x= g*sin(phi) and g_y=- g*cos(phi), where phi is the angle of the slope. Determine launch angle in the rotated coordinate system, call it theta' The kinematic equations are relatively straightforward because in this coordinate sytem the arrows initial and final y-position is zero. Also, the final v y is the negative of the initial

Review | Constants Part A An archer standing on a 15° slope shoots an arrow 20 above the horizontal, as shown in the figure (Figure 1) How far down the slope does the arrow hit if it is shot with a speed of 50 m/s? Hint: There are several strategies to solve this problem. This hint discusses two different methods. Figure Method 1: 1 of 1 The arrow must land on the slope, so write y(x) of the slope as y= tan(phi)x, where phi is angle of the slope and it must be negative. Note that you need to define a coordinate system, which you might assign the origin at the position of the archer, and +x is to the right in graphic. All angles must be with respect to the positive x direction. The slope angle in the picture does not conform to this requirement so you have to determine the slope angle from the information given. That is why you should write -phi for the angle of the slope 200 The distance down the slope is just the sqrt[(displacement in x)^2+(displacement in y)^2]. Strategy: write the usual kinematic equations for x(t) and y(t) with v0 and launch 15° onclo thato ninnn hath aro givon to vou C.htit+n ult) uith tho acuotion tonlohi\, +n ont y)^2]. Strategy: write the usual kinematic equations for x(t) and y(t) with v0 and launch angle theta since both are given to you. Subtitute y(t) with the equation tan(phi)x to get two equations for x as a function of t. From those two equations, solve for time of flight, t of the arrow. Once t is known, you can solve for displacement in x from the usual kinematic equation x(t) and you obtain the displacement in y from the usual kinematic equation for the y-component. Method 2: If you use a rotated coordinate system, so that +x lies in the down direction of the slope, then x is the distance. However in this coordinate system, gravitational acceleration has both x and y components, so this method requires that you solve the kinematic equations for acceleration in both y and x. Determine and x and y components for gravitational acceleration, call them g_x and g_y. You should be able to use geometry to determine that g_x= gsin(phi) and g_y=- g*cos(phi), where phi is the angle of the slope. Determine launch angle in the rotated coordinate system, call it theta' The kinematic equations are relatively straightforward because in this coordinate sytem the arrows initial and final y-position is zero. Also, the final v y is the negative of the initial