The projectile launcher shown below will give the object on the right an inital horizontal speed of 14.3 m/s. While the other object will be dropped with no initial speed. The objects are initially 104 cm above the ground and separated by 142 cm. What will be the difference in the landing locations of the two objects? Could you explain step by step?

To find the difference in the landing locations of the two objects, we need to calculate the horizontal distance traveled by each object.

Let's calculate the time it takes for the two objects to reach the ground using the equation for vertical motion under constant acceleration:

1. For the dropped object:
The initial vertical velocity is 0 m/s, and the acceleration due to gravity is approximately 9.8 m/s^2.
The equation for vertical distance traveled is d = (1/2)gt^2, where d is the distance traveled, g is the acceleration due to gravity, and t is the time.
The vertical distance traveled by the dropped object is 104 cm = 1.04 m.
Substituting these values into the equation, we have:
1.04 m = (1/2)(9.8 m/s^2)(t^2)
Rearranging the equation, we get:
t^2 = (2 * 1.04 m) / 9.8 m/s^2
t^2 = 0.2116 s^2
Taking the square root of both sides, we find:
t ≈ 0.46 s

2. For the projectile launched object:
The initial vertical velocity is also 0 m/s, and the acceleration due to gravity remains 9.8 m/s^2.
The vertical distance is the same (104 cm = 1.04 m).
We can use the same equation as before for the vertical distance traveled:
1.04 m = (1/2)(9.8 m/s^2)(t^2)
Rearranging the equation, we have:
t^2 = (2 * 1.04 m) / 9.8 m/s^2
t^2 = 0.2116 s^2
Taking the square root of both sides, we find:
t ≈ 0.46 s

Now, we can calculate the horizontal distance traveled by each object using the equation d = vt, where d is the distance traveled, v is the initial horizontal velocity, and t is the time:

1. For the dropped object:
The initial horizontal velocity is 0 m/s.
The time calculated earlier is t ≈ 0.46 s.
Therefore, the horizontal distance traveled by the dropped object is:
d = (0 m/s)(0.46 s) = 0 m

2. For the projectile launched object:
The initial horizontal velocity is 14.3 m/s.
The time calculated earlier is t ≈ 0.46 s.
Therefore, the horizontal distance traveled by the projectile launched object is:
d = (14.3 m/s)(0.46 s) ≈ 6.588 m

Finally, to find the difference in the landing locations, we subtract the two distances:

Difference in landing locations = 6.588 m - 0 m = 6.588 m.

Therefore, the difference in the landing locations of the two objects is approximately 6.588 meters.

To find the difference in the landing locations of the two objects, we need to calculate the horizontal distances traveled by each object.

1. First, let's consider the vertically launched object. Since there is no initial vertical speed, it will experience free-fall motion. We can use the formula for free-fall motion to calculate the time it takes for the object to reach the ground:
h = (1/2)gt², where h is the vertical displacement (104 cm) and g is the acceleration due to gravity (9.8 m/s²).

Converting the initial height to meters: 104 cm = 1.04 m.
Rearranging the formula and solving for t:
t = √(2h/g)

Plugging in the values:
t = √(2 * 1.04 m / 9.8 m/s²)
t ≈ √(0.2122)
t ≈ 0.46 s (rounded to two decimal places)

2. Now, let's calculate the horizontal distance traveled by the vertically launched object. We can use the formula:
d = vt, where d is the horizontal distance, v is the horizontal velocity, and t is the time calculated in step 1.

Plugging in the values:
d = 14.3 m/s * 0.46 s
d ≈ 6.58 m (rounded to two decimal places)

3. Finally, let's calculate the difference in landing locations between the two objects. The horizontally launched object will travel horizontally for a distance of 6.58 meters, while the vertically launched object will fall straight to the ground without any horizontal displacement. Since the objects were initially separated by 142 cm (or 1.42 meters), the difference in landing locations will be:
Difference = 6.58 m - 1.42 m
Difference ≈ 5.16 m (rounded to two decimal places)

Therefore, the difference in landing locations of the two objects will be approximately 5.16 meters.

did this already, but you still do not say whether the gun is pointed toward or away from the dropped object.

Anyway, find how long it takes to drop 104cm (value of t)
The horizontal distance covered by the launched object is just 14.3t meters