Question

Amy and bob are in an archery competition. the goal is to hit a target taht is on a wall 50m away from them horizontally. the target is 21m vertically above the ground, and both competitors hold their bows 1m above the ground, so the vertical displacement is 20m. they both use the same bow, which shoots arrows at a speed of 35 m/s at the moment of release.

A) Bob decides to aim directly at the target. what angle has bob chosen?

B) Bob shoots and misses. He forgot to account for the effect of gravity when he aimed; he failed to aim above the target, so his arrow ends up below the target. Exactly how far below the target does Bob's arrow hit the wall?

C)Amy knows what she must aim above the target to make the shot. In fact, there are actually two angles that will cause her to hit the target. What are they?

D)Roughly sketch the path for all three arrows: Bob's miss and Amy's two possible hits.

Answer 1

PROJECTILE

along horizontal
________________

initial velocity vox = v0*costheta

acceleration ax = 0

initial position = xo = 0

final position = x

displacement = x - x0

from equation of motion

x - x0 = v0x*T+ 0.5*ax*T^2

x - x0 = vo*costheta*T

T = (x - x0)/(vo*costheta)......(1)

along vertical
______________

initial velocity v0y = vo*sintheta

acceleration ay = -g = -9.8 m/s^2

initial position = y0

final position = y

from equation of motion

y-y0 = v0y*T + 0.5*ay*T^2 .........(2)

using 1 in 2

y-y0 = (vo*sin(theta)*(x-x0))/(vo*cos(theta)) - (0.5*g*(x-x0)^2)/(vo^2*(cos(theta))^2)

y - y0 = (x - x0)*tantheta - (0.5*g*(x-x0)^2)/(vo^2*(cos(theta))^2)

(A)

tantheta = 20/50 = 22 degrees

(B)

yo = 1m

y = ?

y - 1 = 50*tan22 - (0.5*9.81*50^2)/(35^2*(cos22)^2)

y = 9.56 m

(C)

for y = 21 and x- xo = 50

21 - 1 = 50*tantheta - (0.5*9.81*50^2)/(35^2*(costheta)^2)

theta = 35 and 55

(D)