Aprojectile is fired from level ground with an initial speed of 24.3 m/s at a launch angle of 32 degrees relative to the horizontal direction. where does the projectile land if you ignore air resistance? (assume 3 significant digits).

The projectile lands at a distance of 54.2 m from the launching point.

Explanation:

The projectile´s position is described by the following position vector:

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.

t = time.

α = launching angle.

y0 = initial vertical position.

g = acceleration due to gravity (-9.8 m/s² considering the upaward direction as positive).

When the projectile lands, its vector position is "r final" (see figure). Notice that the y-component of that vector is 0 m if we place the origin of the frame of reference at the launching point. Then, using the equation for the vertical position of the projectile:

y = y0 + v0 · t · sin α + 1/2 · g · t²

Notice that y0 = 0 because the origin of the frame of reference is located at the launching point. Then, let´s write the equation when y = 0 and solve for the time:

0 = 24,3 m/s · t · sin 32° - 1/2 · 9.8 m/s² · t²

0= t ( 24.3 m/s · sin 32° - 4.9 m/s² · t)       Solution 1 t = 0 (the initial position)

0 = 24.3 m/s · sin 32° - 4.9 m/s² · t

- 24.3 m/s · sin 32 / - 4.9 m/s² = t

t = 2.63 s

Now, we can calculate the x- component of the vector r final that is the horizontal distance covered by the projectile:

x = x0 + v0 · t · cos α

x = 0 m + 24.3 m/s · 2.63 s · cos 32°

x = 54.2 m

The projectile lands at a distance of 54.2 m from the launching point.

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