An arrow is fired with a velocity of 93 m/s at an angle of 18 degrees to the horizontal. How far does the arrow travel?

Question

Projectile motion at angles.

Answer 1

The only force acting on the arrow is assumed to be gravity, which acts downward in the y direction. There are no forces in the x direction, so we can write out the distance that the arrow travels, d, as:

v_x = d / t -> d = v_x * t

Where v_x is the velocity in the x direction and t is the time that the arrow is in the air. Notice that this is only true because there are no forces in the x direction, so the velocity of the arrow in the x direction is constant. Now we know the total velocity v and the angle from the horizontal, so v_x = v * cos(theta) = 93 m/s * cos(18 degrees). The only variable we are missing is t, the time that the arrow is in the air.

We can calculate the time the arrow is in the air as y = y₀ + v_0y * t - 0.5 * g * t² where t is the time the arrow is in the air, v_0y = v * sin(theta) = 93 m/s * sin(18 degrees) is the initial velocity of the arrow in the y direction, and y = y₀ = 0 are the initial and final y values of the arrow. Since we want to find the time the arrow is in the air (so from the ground where it is fired until it lands back at the ground) we set both to 0. We can then solve this equation for t, so 0 = 0 + 93 m/s * sin(18 degrees) * t - 0.5 * g * t² and get t = 2 * 93 m/s * sin(18 degrees) / g.

Plugging our equation for t into our original equation for d that we are solving for, we find that d is around 518.8 meters, or 520 meters rounding for significant figures.

Answer 2

We need to use Second Newton's Law (Fondamental Relation of Dynamics): F = m.a (with F and a which are vectors).

The only force is the weight P = m.g.

So: m.g = m.a.

And: a = g.

With the x-axis for horizontal (right) and the y-axis for vertical (up), we have: a_x = 0 and a_y = -g.

We know that a = dv/dt = g so:

a_x = dv_x/dt = 0.

a_y = dv_y/dt = -g.

And:

v_x = constant = v₀.cos(alpha).

v_y = -g.t + constant = -g.t + v₀.sin(alpha)

We know that v = dOM/dt so:

v_x = v₀.cos(alpha) = dx/dt.

v_y = -g.t + v₀.sin(alpha) = dy/dt.

And:

x = v₀.cos(alpha).t + constant = v₀.cos(alpha).t

y = -0.5.g.t² + v₀.sin(alpha).t + constant = -0.5.g.t² + v₀.sin(alpha).t

To know how far does the arrow travel, we write y = 0 so:

y = -0.5.g.t² + v₀.sin(alpha).t = 0 so t = 0 or t = (v₀.sin(alpha))/(0.5.g)

And: x_max = v₀.cos(alpha)((v₀.sin(alpha))/(0.5.g)) = 2.v₀²cos(alpha).sin(alpha)/g.

So: x_max = 2.93².cos(18).sin(18)/9.81 = 519 m so x_max = 520 m (with 2 sig. fig).

Note: You must do that in 5 min, why not without the demonstration of:

x = v₀.cos(alpha).t.

y = -0.5.g.t² + v₀.sin(alpha).t.