1.You are stuck on a cliff and want to know how far to the bottom. You drop a rock and time how long it takes to hit the ground. If the rock takes 10 seconds to hit the ground how high is the cliff? How fast was the rock going to impact?

2. You throw a plate off your roof with a speed of 10.0 m/ s. 4 seconds later it hits the ground? How far?
3. You throw a baseball straight up at 40 m/ s and it reaches its peak after 4 seconds. What is its speed when it falls back into your hand? What is it displacement? What distance has it traveled?

hf=hi+vi*t+1/2 gt^2

g=-9.8m/s^2

in the first, hf=0,hi=?, vi=0
in the second, vi=-10,hf=0, hi=?

in the third, the velocity at the top is zero.

Vf=vi+gt
vf=40-9.8(8)
displacement =0 if it returns to your hand.

distance? twich the height..
hf=hi+40*t+1/2 g t^2 g=-9.8m/s^2

To solve these problems, we can use the laws of motion and some basic equations of kinematics. Let's go through each question step by step:

1. Stuck on a cliff:
To determine the height of the cliff, we can use the equation for free fall motion:

h = (1/2)gt^2

where h is the height, g is the acceleration due to gravity (approximately 9.8 m/s^2), and t is the time it takes for the rock to hit the ground (given as 10 seconds).

Substituting the values into the equation:

h = (1/2)(9.8 m/s^2)(10 s)^2
h = 490 m

Therefore, the height of the cliff is 490 meters.

To calculate the rock's impact velocity, we can use the equation:

v = gt

where v is the velocity and g is the acceleration due to gravity. Since the rock is falling freely, the downward acceleration is equal to the acceleration due to gravity.

Substituting the values:

v = (9.8 m/s^2)(10 s)
v = 98 m/s

Therefore, the rock was going at a speed of 98 m/s when it impacted.

2. Plate off the roof:
To find the distance the plate traveled, we can use the equation of motion:

s = ut + (1/2)at^2

where s is the distance, u is the initial velocity, t is the time, and a is the acceleration.

Given: u = 10.0 m/s, t = 4 s, and a = 9.8 m/s^2 (acceleration due to gravity).

Substituting the values:

s = (10.0 m/s)(4 s) + (1/2)(9.8 m/s^2)(4 s)^2
s = 40 m + 78.4 m
s = 118.4 m

Therefore, the plate traveled a distance of 118.4 meters.

3. Baseball thrown straight up:
When the baseball reaches its peak, its velocity becomes 0 m/s. At this point, it starts to fall back down due to the acceleration due to gravity. The final velocity when it falls back into your hand is the opposite of the initial velocity.

Given: initial velocity (upwards) = 40 m/s, time to reach peak = 4 s, and acceleration due to gravity = 9.8 m/s^2.

To find the final velocity when it falls back down, we can use the equation:

v = u + at

where v is the final velocity, u is the initial velocity, a is the acceleration, and t is the time.

Substituting the values:

v = 40 m/s + (-9.8 m/s^2)(4 s)
v = 40 m/s - 39.2 m/s
v = 0.8 m/s

Therefore, the speed of the baseball when it falls back into your hand is 0.8 m/s. This indicates that the ball is moving downwards.

To find the displacement, we can use the equation:

s = ut + (1/2)at^2

Since the initial velocity is upwards, we take the initial velocity as positive and the displacement will also be positive.

s = (40 m/s)(4 s) + (1/2)(-9.8 m/s^2)(4 s)^2
s = 160 m - 78.4 m
s = 81.6 m

Therefore, the displacement of the baseball is 81.6 meters.

To find the total distance traveled by the baseball, we add the distance traveled upwards to the displacement:

distance = 81.6 m + 40 m
distance = 121.6 m

Therefore, the baseball has traveled a total distance of 121.6 meters.