a person in a car traveling at 18meters/s drops a cherry pit out of the window 1meter above the ground.
A. how far horrizontal will the pit hit hit ground from the initial dropping point ?
B. if the car continues to travel at the same speed where will the car be in relation to the cherry pit?
B. First because easy.
The car and the pit both travel with the same horizontal speed component until the pit hits the dust
so
the pit hits right under the window.
Now A
How long to drop one meter?
d = (1/2)a t^2
1 = 4.9 t^2
t = .452 seconds
How far will it go at 18 m/s for .452 seconds?
18*.452 = 8.13 meters
thank you so much
You are welcome. Remember whenever you bomb something turn the airplane immediately after releasing the bomb. Otherwise the bomb is likely to hit right under the plane.
Haha ok i will for sure remember that
To solve both parts of the problem, we can use the equations of motion. Let's break it down step by step:
A. How far horizontally will the pit hit the ground from the initial dropping point?
To find the horizontal distance traveled by the cherry pit, we need to determine the time it takes for the pit to fall. We'll use the equation:
h = 1/2 * g * t^2
Where:
h = vertical distance traveled (in this case, 1 meter)
g = acceleration due to gravity (approximately 9.8 m/s^2)
t = time
Rearranging the equation to solve for t, we get:
t^2 = (2h) / g
t = sqrt((2h) / g)
Now we have the time it takes for the cherry pit to fall. We can multiply that by the horizontal velocity (18 m/s) to find the horizontal distance traveled:
horizontal distance = time * velocity
Substituting the values into the equation:
horizontal distance = (sqrt((2 * 1) / 9.8)) * 18
After performing the calculations, we find that the pit will hit the ground approximately 3.43 meters away horizontally from the initial dropping point.
B. If the car continues to travel at the same speed, where will the car be in relation to the cherry pit?
Since the car and the cherry pit are both traveling at the same velocity (18 m/s) horizontally, the distance between them will remain constant. Therefore, the car will stay approximately 3.43 meters away from the cherry pit throughout its fall.
Note: Keep in mind that this calculation assumes there are no external factors such as air resistance or any changes in the car's velocity.