A person stands on a bridge that is 100 feet above a river.

If she drops a pebble how fast is it moving after 2 seconds?
How long does it take the pebble to reach the river below?
She has another pebble that she tosses up with initial speed of 8 feet per second. It goes up, then starts falling down to the river below. By the time it reaches the river what is its speed?

Calculus? Hardly.

vf=vi+gt for the firstquestion

second question: h=vi*t-4.9t^2 Vi=0, solve for t

third question
hf=hi+vi*t-4.9t^2
0=100+8t-4.9t^2
solve for time t, then vf=vi-9.8t

but there is an easier method for the last:
vf^2=vi^2+2ad where a=-9.8, d=-100

To answer these questions, we can use the equations of motion and some basic principles of physics.

For the first question, "How fast is the pebble moving after 2 seconds?", we need to consider the acceleration due to gravity. According to the equations of motion, the vertical velocity of an object in free fall increases by 9.8 meters per second per second (or 32.2 feet per second per second) due to gravity. Since the pebble is dropped without any initial upward velocity, its velocity will increase solely due to gravity.

To find the velocity after 2 seconds, we can use the equation:

v = u + at

where:
v is the final velocity,
u is the initial velocity (which is 0 in this case),
a is the acceleration due to gravity (approximately 32.2 feet per second per second),
t is the time.

Plugging in the values, we have:

v = 0 + (32.2 ft/s^2) * 2 s
v = 64.4 ft/s

Therefore, after 2 seconds, the pebble will be moving downward with a velocity of 64.4 feet per second.

For the second question, "How long does it take the pebble to reach the river below?", we need to find the time it takes for the pebble to travel a distance of 100 feet.

We can use the equation:

s = ut + (1/2) a t^2

where:
s is the distance,
u is the initial velocity (which is 0 in this case),
a is the acceleration due to gravity (approximately 32.2 feet per second per second),
t is the time.

In this case, s is 100 feet, so we have:

100 = 0 * t + (1/2) * 32.2 * t^2

Rearranging the equation, we get:

16.1 t^2 = 100

Dividing both sides by 16.1, we get:

t^2 = 100 / 16.1

Taking the square root of both sides:

t ≈ √6.21

t ≈ 2.49 seconds

Therefore, it will take approximately 2.49 seconds for the pebble to reach the river below.

For the third question, "By the time it reaches the river, what is the speed of the pebble?", we need to find the speed at the moment the pebble reaches the river.

We can use the same equation as in the first question:

v = u + at

However, in this case, the initial velocity is upward with a value of 8 ft/s, and the acceleration due to gravity is acting downward, so it has a negative value.

Plugging in the values, we have:

v = 8 ft/s + (-32.2 ft/s^2) * t

Using the time calculated in the second question (t ≈ 2.49 seconds), we can find the final velocity:

v = 8 ft/s + (-32.2 ft/s^2) * 2.49 s
v ≈ -70.98 ft/s

Since the velocity is negative, it means the pebble is moving downward. Therefore, the speed of the pebble when it reaches the river is approximately 70.98 feet per second.