Horizontal Range Formula

A projectile is an object that we give an initial velocity, and the gravity acts on it. A projectile’s horizontal range is the distance along the horizontal plane. Moreover, it would travel before it reaches the same vertical position as it started from. After that, the horizontal range is depending upon the initial velocity \(V_{0}\), the launch angle \(\theta\), and the acceleration occurring due to the gravity. The horizontal range’s unit is meters (m). Learn horizontal range formula here.

Projectile Motion Formula 

A projectile is an object that is in flight after we throw it or project it. In projectile motion, the only acceleration acting is in the direction and the direction is in the vertical direction. n other words, its the acceleration due to gravity (g). The equations of the motion are applicable separately in X-Axis and the Y-Axis for finding the unknown parameters.

Horizontal Range = \(\frac{horizontal range = (initial velocity)^{2}(sine of 2 X launch angle)}{2(acceleration due to gravity)}\)
R = \(\frac{v_{0}^{2}sin 2\theta}{g}\)

Here:

R = horizontal range (m)
\(V_{0}\) = initial velocity (m/s)
G = acceleration due to gravity (\(9.8m/s^{2}\))
\(\theta\) = angle of the initial velocity from the horizontal plane (radians or degree)

Derivation of the Horizontal Range Formula

Most of the basic physics textbooks talk on the topic of horizontal range of the Projectile motion. Therefore, we derive it using the kinematics equations:

\(a_{x}\) = 0
\(v_{x}\) = \(v_{0x}\)
\(\triangle x\) = \(v_{0x}t\)
\(a_{y}\) = -g
\(v_{y}\) = \(v_{0y}\) – gt
\(\triangle y\) = \(v_{0y}t\) – \(\frac{1}{2}gt^{2}\)

Where,

\(v_{0x}\) = \(v_{0}cos\theta\)
\(v_{0y}\) = \(v_{0}sin\theta\)

Suppose a projectile is thrown from the level of the ground, thus, the range is the distance between the launch point and landing point, where the projectile is hitting the ground. Further, when the projectile returns to the ground, the vertical displacement is zero, thus we have:

0 = \(v_{0}sin\theta\)t – \(\frac{1}{2}gt^{2}\)

Solving for t, we have

t = 0, \(\frac{2_{v0}sin\theta}{g}\)

The first solution provides the time when the projectile is thrown and the second one is the time when it hits the ground. Plugging in the second solution into the displacement equation and using 2 sin \(\theta\) cos \(\theta\) = sin(2\(\theta\)), we have

R = \(\Delta x(t= 2_{v_0 sin\theta/g})\) = \(\frac{v_{0}^{2}}{g}sin(2\theta)\)

Solved Examples on Horizontal Range Formula

1) A motorcyclist has set up a bike stunt with a ramp at the edge of a gorge 75.0 m broad. The ramp is inclined by 53.1° from the horizontal plane. She plans to take off the bike from the ramp at a velocity of 28.0 m/s. At that velocity, what will be the horizontal range, and will she be a success in going to the other side of the gorge?

Solution: We can get the horizontal range of the motorcyclist by using the formula:

R = \(\frac{v{_{0}}^{2}sin2\theta}{g}\)
R = \(\frac{(28.0m/s)^{2}sin2(53.1°)}{9.80 m/s^{2}}\)
R = \(\frac{(28.0m/s)^{2}sin(53.1°)}{106.2 m/s^{2}}\)
R \(\cong\) \(\frac{(28.0m/s)^{2}(0.960)}{9.80 m/s^{2}}\)
R \(\cong\) \(\frac{(784m^{2}/s^{2})(0.960)}{9.80 m/s^{2}}\)
R \(\cong\) 76.8 m

The horizontal range of the motorcyclist will be 76.8 m if she takes off the bike from the ramp at 28.0 m/s. This is a little bit greater than the 75.0 m width of the gorge, so she will make it to the different side.

2) A pirate fired one of the cannons of the ship to test its range. It was set at an angle of 18.5°. The pirate watched the cannonball and noted that it hit the water at a distance of 800 m away. What was the cannon ball’s velocity when it was leaving the cannon?

Solution: The cannon ball’s velocity is found by rearranging the horizontal range formula:

R = \(\frac{v{_{0}}^{2}sin2\theta}{g}\) \(\rightarrow\) \(v{_{0}}^{2}\) = \(\frac{gR}{sin2\theta}\)
\(v{_{0}}^{2}\) = \(\frac{gR}{sin2\theta}\) \(\rightarrow\) \(v_{0}\) = \(\sqrt{\frac{gR}{sin2\theta}}\)
\(v_{0}\) = \(\sqrt{\frac{(9.80m/s^{2})(800m)}{sin2(18.5°)}}\)
\(v_{0}\) = \(\sqrt{\frac{(9.80m/s^{2})(800m)}{sin(37.0°)}}\)
\(v_{0}\) \(\cong\) \(\sqrt{\frac{(9.80m/s^{2})(800m)}{0.6018}}\)
\(v_{0}\) \(\cong\) \(\sqrt{13028m^{2}/s^{2}}\)
\(v_{0}\) \(\cong\) 114 m/s

The initial velocity of the cannon ball is approximately 114 m/s.