An archer on ground that has a constant upward slope of 30.0 ? aims at a target

Question

An archer on ground that has a constant upward slope of 30.0 ? aims at a target 60.0 m farther up the incline. The arrow in the bow and the bull's-eye at the center of the target are each 1.50 m above the ground. The initial velocity of the arrow just after it leaves the bow has magnitude 32.0 m/s.

A) At what angle above the horizontal should the archer aim to hit the bull's-eye? If there are two such angles, calculate the smaller of the two. You might have to solve the equation for the angle by iteration

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


====================================================================================

Related Questions

Navigate

• Browse (All)
• Browse A
• Subjects

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