tad drops a cherry pit out the car window 1 m above the ground while traveling down the road at 18 m/s.

a.how far, horizontally, from the initial dropping point will the pit hit the ground?
b. if the car continues to travel at the same speed, where will the car be in relation to the pit when it lands?

b is really easy. If there is no air friction, both the car and the pit continue at the same horizontal velocity and the pit lands right under the window.

Now part A
how long to fall 1 meter?
0 = 1 + 0 - 4.9 t^2
t^2 = 1/4.9
t = sqrt (1/4.9)
t = .452 s
how far does it go in .452 s at 18 m/s?
18*.452 = 8.13 meters

Physics man approves

thanks

WHy is 1-0 in the eguation?

To solve this problem, we can use the equations of motion to find the distance traveled by the cherry pit horizontally and the position of the car when the pit hits the ground.

a. First, let's find the time it takes for the cherry pit to hit the ground. We'll use the equation:

h = (1/2) * g * t^2

Where h is the initial height of the pit above the ground (1 m), g is the acceleration due to gravity (approximately 9.8 m/s^2), and t is the time it takes the pit to hit the ground.

Plugging in the values, we get:

1 = (1/2) * 9.8 * t^2

Solving for t, we find:

t^2 = (2 * 1) / 9.8
t^2 = 0.2041
t ≈ 0.452 s

Next, we can find the horizontal distance traveled by the cherry pit using the equation:

d = v * t

Where d is the horizontal distance, v is the velocity of the car (18 m/s), and t is the time calculated above.

Plugging in the values, we get:

d = 18 * 0.452
d ≈ 8.14 m

Therefore, the cherry pit will hit the ground approximately 8.14 meters horizontally from the initial dropping point.

b. Since the car continues to travel at the same speed, it will also travel a distance of 8.14 meters during the time it takes the pit to hit the ground. Therefore, the car will be at the same horizontal position as the pit when it lands.