The projectile launcher shown below will give the object on the right an inital horizontal speed of 8.0 m/s. While the other object will be dropped with no initial speed. The objects are initially 137 cm above the ground and separated by 142 cm. What will be the difference in the landing locations of the two objects?
the horizontal speed is constant.
distance = speed * time
So, you need to know
(a) how long does it take to fall 1.37m? (s = 1/2 at^2)
(b) Is the launcher pointed toward or away from the dropped object? No way to tell from what you said, and no diagram to check.
To determine the difference in landing locations of the two objects, we need to calculate the horizontal distance traveled by each object.
The horizontal distance traveled by an object can be calculated using the formula:
distance = velocity × time
For the object launched from the projectile launcher, the initial horizontal speed is given as 8.0 m/s.
For the object dropped with no initial speed, the initial horizontal speed is 0 m/s.
Let's calculate the time taken for each object to reach the ground:
For the object launched from the projectile launcher:
First, we need to find the time it takes for the object to fall from an initial height of 137 cm. Using the equation:
distance = (1/2) × g × t^2
where g is the acceleration due to gravity (approximately 9.8 m/s^2), and t is the time.
Simplifying the equation, we have:
137 cm = (1/2) × 9.8 m/s^2 × t^2
Converting 137 cm to meters, we have:
1.37 m = (1/2) × 9.8 m/s^2 × t^2
Rearranging the equation to solve for t^2, we get:
t^2 = (2 × 1.37 m) / 9.8 m/s^2
t^2 = 0.2795918 s^2
Taking the square root of both sides, we find:
t = 0.5289635 s (approximately)
Now, we can calculate the horizontal distance traveled by the launched object:
distance = velocity × time
distance = 8.0 m/s × 0.5289635 s
distance ≈ 4.232 m
For the object that is dropped:
Since there is no initial horizontal speed, the time it takes for the object to reach the ground is the same as the time calculated above, t ≈ 0.5289635 s.
The horizontal distance traveled by the dropped object is:
distance = velocity × time
distance = 0 m/s × 0.5289635 s
distance = 0 m
Therefore, the difference in landing locations of the two objects is:
4.232 m - 0 m = 4.232 m
The objects will have a difference of 4.232 meters in their landing locations.
To find the difference in the landing locations of the two objects, we need to determine the horizontal distance traveled by each object.
We can start by calculating the time it takes for the objects to hit the ground.
For the projectile launcher, we need to find the time it takes for the object to travel 142 cm horizontally at an initial speed of 8.0 m/s. We can use the equation:
distance = speed × time
142 cm = 8.0 m/s × time
To solve for time, we need to convert 142 cm to meters:
142 cm = 1.42 m
1.42 m = 8.0 m/s × time
Now solve for time:
time = 1.42 m / 8.0 m/s
time ≈ 0.18 seconds
Therefore, the object launched by the projectile launcher will take approximately 0.18 seconds to hit the ground.
For the dropped object, we can calculate the time it takes for the object to fall from a height of 137 cm (converted to meters) using the equation:
distance = 1/2 × acceleration due to gravity × time^2
137 cm = 1/2 × 9.8 m/s^2 × time^2
Convert 137 cm to meters:
137 cm = 1.37 m
1.37 m = 1/2 × 9.8 m/s^2 × time^2
Now solve for time:
time = √((2 × distance) / acceleration due to gravity)
time = √((2 × 1.37 m) / 9.8 m/s^2)
time ≈ 0.167 seconds
Therefore, the dropped object will take approximately 0.167 seconds to hit the ground.
Now, we can find the horizontal distance traveled by each object:
For the projectile launcher:
distance = speed × time
distance = 8.0 m/s × 0.18 s
distance ≈ 1.44 meters
For the dropped object:
distance = 0 s (since it has no initial horizontal speed)
Therefore, the difference in the landing locations of the two objects is approximately 1.44 meters.