A hungry hunter of early middle ages who had a bow and an arrow noticed a deer t

Question

A hungry hunter of early middle ages who had a bow and an arrow noticed a deer that was eating grass at the distance of 40 meters from him on a horizontal empty field. The hunter was lucky to kill the deer. Please determine how he directed his arrow at the initial moment of time it is known that: The initial velocity of the arrow was 50 (m/s). The height of the hunter was 1.7 (m). The height of the deer was 0.6 (m). The air resistance can be neglected. The hunter used the trajectory for which the time of the flight of the arrow was the shortest one. The acceleration due to gravity is equal to 9.81 (m s^-2). The student solved the problem and found that the angle between the arrow and the horizontal line at the initial moment of time, measured in degrees and rounded-off to three decimal digits was as follows: (your numerical answer must be written here)

Explanation / Answer

PROJECTILE

along horizontal

initial velocity vox = vo*costheta

ax = 0

from equation of motion

x-Xo = vox*T+ 0.5*ax*T^2

x-X0 = vo*costheta*T

T = (x-X0)/(vo*costheta)......(1)

along vertical

voy = vo*sintheta

acceleration ay = -Eq/m = 6.4*10^5*1.6*10^-19/(1.67*10^-27) = -6.13*10^13 m/s^2

from equation of motion

y-y0 = voy*T + 0.5*ay*T^2

y-y0 = (vo*sintheta_0*(x-x0))/(vo*costheta_0) + (0.5*ay*(x-x0)^2)/(vo^2*(costheta_0)^2)

y-y0 = (x-X0)*tantheta_0 + ((0.5*ay*(x-X0)^2)/(vo^2*(costheta_0)^2))

y0 = 1.7 m

y = 0.6 m

x - x0 = 40 m

theta = ?

v0 = 50 m/s

costheta = x

tantheta = sintheta/costheta = sqrt(1-x^2)/x

0.6-1.7 = (40*sqrt(1-x^2)/x) - (0.5*9.81*40^2/(50^2*x^2))

x = costheta = 0.9986925848856

theta = 2.93 degrees

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