Answer:
The horizontal distance traveled by the shot is 21.4 m
Explanation:
The position of the shot is given by the position vector r
r = (x0 + v0 · t · cos α, y0 + v0 · t · sin α + 1/2 · g · t²)
Where:
r = position vector at time t
x0 = initial horizontal position
v0 = initial velocity
t = time
α = throwing angle
y0 = initial vertical position
g = acceleration due to gravity (-9.8 m/s² considering the upward direction as positive).
Please, see the attached figure for a description of the situation. Notice that when the shot hits the ground, the y-component of the position vector of the shot is -2.20 m (origin of the frame of reference located at the throwing point). Using this data, we can obtain the time at which the shot hits the ground and calculate the x-component of "r final", which is the horizontal distance traveled by the shot. Then:
y = y0 + v0 · t · sin α + 1/2 · g · t²
-2.20 m = 0m + 14.0 m/s · t · sin 35° - 1/2 · 9.8 m/s² · t²
0 = -4.9 m/s² · t² + 14.0 m/s · t · sin 35° + 2.20 m
Solving the quadratic equation:
t = 1.87 s
Now, with this time, we can calculate the x-component of "r final"
x = x0 + v0 · t · cos α
x = 0 m + 14.0 m/s · 1.87 s · cos35°
x = 21.4 m
The horizontal distance traveled by the shot is 21.4 m
