Calculate Horizontal Range of Projectile - Unit Vector Velocity

[George3]

Homework Statement

Exactly 2.9 seconds after a projectile is fired it has the velocity (9.0i +4.5j) m/s.
What is the horizontal range of the projectile.

Homework Equations

The Attempt at a Solution

Range = [((9.0i^2 + 4.5j^2)^1/2)^2 x sin2(arctan(4.5/9))]/ 9.8 m/s/s
Range = 8.3 m but this is incorrect.

 

[D H]

Of course that's not correct. Look at the units, for one thing. For another, you aren't accounting for that 2.9 seconds.

 

[tomcue]

I would first check the units of the given velocity and make sure they are consistent. Since the units are in meters per second, we can proceed with the calculation. However, the given velocity is not a unit vector, as it has magnitudes in both the i and j directions. We need to convert it to a unit vector by dividing each component by the magnitude of the vector. This will give us a unit vector velocity of (0.8944i + 0.4472j) m/s.

Next, we can use the formula for horizontal range: Range = (v^2/g) * sin2θ, where v is the initial velocity, g is the acceleration due to gravity, and θ is the launch angle. Since we are dealing with a horizontal range, the launch angle is 0 degrees. Plugging in the given values, we get Range = (8.3 m/s)^2 * sin0 = 0 m. This result is not surprising, as the projectile is fired horizontally and will not travel any distance in the horizontal direction in 2.9 seconds.

In conclusion, the horizontal range of the projectile in this scenario is 0 meters. This result aligns with our intuition, as the given velocity is purely horizontal and there is no vertical component to the motion.