An archer on ground that has a constant upward slope of 30 degreesaims at a targ

Question

An archer on ground that has a constant upward slope of 30 degreesaims at a target 60.0 m farther up the incline. The arrow in thebow and the bull's eye at the center of the target are each 1.50 mabove ground. The initial velocity of the arrow just after itleaves the bow has magnitude 32.0 m/s. At what angle above thehorizontal should the archer aim to hit the bull's-eye?
If there are twosuch angles, calculate the smaller of the two. You might have tosolve the equation for the angle by angle iteration, that is, bytrial and error.
How does the angle compare to that required when the ground islevel, with zero slope?
b)repeat above for ground that has downward slope of 30degrees.
i have no idea withthis one please help me toanswer this question thanks
If there are twosuch angles, calculate the smaller of the two. You might have tosolve the equation for the angle by angle iteration, that is, bytrial and error.
How does the angle compare to that required when the ground islevel, with zero slope?
b)repeat above for ground that has downward slope of 30degrees.
i have no idea withthis one please help me toanswer this question thanks If there are twosuch angles, calculate the smaller of the two. You might have tosolve the equation for the angle by angle iteration, that is, bytrial and error.
How does the angle compare to that required when the ground islevel, with zero slope?
b)repeat above for ground that has downward slope of 30degrees.
i have no idea withthis one please help me toanswer this question thanks

Explanation / Answer

Consider the triangle formed by the archer, the target and the horizontal. This is just a right triangle with a 30 degree angle at the archer and this can be used to find horizontal and vertical distances between the points. Use the formula for height as a function of time (using the initial point as zero). v0 = inititial vertical velocity s = v0*t - (1/2)*g*t^2 Solve for t^2: t^2 = (2*v0/g)*t - 2*s/g Now find an expression for v0. v0 = v*sin(A) v1 = horizontal velocity = v*cos(A) and v = v1/cos(A) But distance is velocity times time so: D = v1*t and v1 = D/t So: v = D/[t*cos(A)] Giving: v0 = v*sin(A) = (D/t)*tan(A) Now use this in the equation for t^2. t^2 = (2*D/g)*tan(A) - 2*s/g Also use the horizontal motion to get an expression for t: t = D/v1 = D/[v*cos(A)] Square this and equate the two expressions for t^2. D^2/[v^2*cos^2(A)] = (2*D/g)*tan(A) - 2*s/g We should use some real values now. D = 50*cos(30) = 30*SQRT(3) s = the vertical separation of the two points = 60*sin(30) = 30 g*45/v^2 = SQRT(3)*sin(A)*cos(A) - cos^2(A) And some trig identities. sin(A)*cos(A) = sin(2A)/2 = SQRT[1 - cos^2(2A)]/2 cos^2(A) = [1 + cos(2A)]/2 g*90/v^2 = SQRT(3)*SQRT[1 - cos^2(2A)] - [1 + cos(2A)] Let M = g*90/v^2 + 1 M + cos(2A) = SQRT(3)*SQRT[1 - cos^2(2A)] M^2 + 2*M*cos(2A) + cos^2(2A) = 3*[1 - cos^2(2A)] 4*cos^2(2A) + 2*M*cos(2A) + M^2 - 3 = 0 And this is just a quadratic in cos(2A) and the solution is the standard: cos(2A) = [-b +/- SQRT(b^2 - 4*a*c)]/(2*a) Where a = 4; b = 2*M and c = M^2 - 3 With M = g*90/v^2 + 1 = 1.8622 We want the smaller angle so we want the +" solution. cos(2A) = -0.149664622 A = 0.8605 radians = 49.3 degrees So the archer the aim at an agnle of 49.3 above the horizontal to hit the target. (B) 60 = v*cos(A)*t and t = 60/[v*cos(A)] v*sin(A) = g*t/2 = (g/2)*60/[v*cos(A)] sin(A)*cos(A) = 30*g/v^2 = sin(2A)/2 sin(2A) = 60*g/v^2 A = 0.30618 radians = 17.543 degrees (C) This is the same as A only the vertical separation is now s = -30. g*45/v^2 = SQRT(3)*sin(A)*cos(A) + cos^2(A) And some trig identities. sin(A)*cos(A) = sin(2A)/2 = SQRT[1 - cos^2(2A)]/2 cos^2(A) = [1 + cos(2A)]/2 g*90/v^2 = SQRT(3)*SQRT[1 - cos^2(2A)] + [1 + cos(2A)] Let M = g*90/v^2 - 1 M - cos(2A) = SQRT(3)*SQRT[1 - cos^2(2A)] M^2 - 2*M*cos(2A) + cos^2(2A) = 3*[1 - cos^2(2A)] 4*cos^2(2A) - 2*M*cos(2A) + M^2 - 3 = 0 Again a quadratic with a = 4; b = -2*M and c = M^2 - 3 With M = -0.137792969 This gives cos(2A) = 0.829519324 And A = -0.2963 radians = -16.975 degrees Any of these can be checked by calculating the time from the horizontal motion and then using that time to calculate the vertical distance and verify that it is correct. For C, for example. The arrow must drop 30 m and travel 20*SQRT(3) m horizontally. V1 = 30.6 m/s and V0 = -9.34 m/s t = D/V1 = 30*SQRT(3)/20.6 = 1.698 s s = V0*t - (1/2)g*t^2 = -9.34*1.698 - (1/2)*9.81*(1.698)^2 = -30 And this is correct

Related Questions

Navigate

• Browse (All)
• Browse A
• Subjects

---

---

Integrity-first tutoring: explanations and feedback only — we do not complete graded work. Learn more.