# recall that the acceleartion due to earth's gravity is 32 ft/sec^2. from ground level, a projectile is fired straight upward with velocity 90 feet per second.

a. what is its velocity after 3 seconds

b. when does it hit the ground

c. when it hits the ground, what's the net distance it has traveled.

d. when it hits the ground, what is the total distance it has traveled

## a) use v = Vo - gt

b ) use equation in part a to find out when v = 0. At that point the thing is as high as it will go. It will spend the same time falling so double the time you found for ascent to find out when it hits the ground.

c) LOL it came right back to where it started if you shot straight up :)

d) we know how long it took to get to the top (half of the final answer in part b)

so how far up did it go?

Y = Yo + Vo t +(1/2) a t^2

Y = 0 + 90 t - 16 t^2 where t is half the total time aloft from part b

double that answer to include the distance it fell.

## simplifing fractions

## alyssa in general factor top and bottom and cancel like factors

for example 12 /16

= 2*2 *3

-----------

2*2 *2*2

so it is really 3/4

## To solve these problems, we can use the equations of motion considering the acceleration due to gravity. Let's break down each question separately.

a. What is its velocity after 3 seconds?

To find the velocity after 3 seconds, we can use the equation:

final velocity (v) = initial velocity (u) + acceleration (a) * time (t)

Given:

u = 90 ft/sec (initial velocity)

a = 32 ft/sec^2 (acceleration due to gravity)

t = 3 sec (time)

Substituting the values into the equation:

v = 90 ft/sec + (32 ft/sec^2 * 3 sec)

v = 90 ft/sec + 96 ft/sec

v = 186 ft/sec

Therefore, the velocity of the projectile after 3 seconds is 186 ft/sec.

b. When does it hit the ground?

To find the time it takes for the projectile to hit the ground, we can use the equation:

time (t) = (final velocity (v) - initial velocity (u)) / acceleration (a)

Given:

v = 0 ft/sec (final velocity, since it hits the ground)

u = 90 ft/sec (initial velocity)

a = 32 ft/sec^2 (acceleration due to gravity)

Substituting the values into the equation:

t = (0 ft/sec - 90 ft/sec) / -32 ft/sec^2

t = -90 ft/sec / -32 ft/sec^2

t = 2.8125 sec

Therefore, it takes approximately 2.81 seconds for the projectile to hit the ground.

c. When it hits the ground, what's the net distance it has traveled?

To find the net distance (displacement) traveled by the projectile, we can use the equation:

distance (s) = initial velocity (u) * time (t) + (1/2) * acceleration (a) * time (t)^2

Given:

u = 90 ft/sec (initial velocity)

a = 32 ft/sec^2 (acceleration due to gravity)

t = 2.8125 sec (time obtained in part b)

Substituting the values into the equation:

s = 90 ft/sec * 2.8125 sec + (1/2) * 32 ft/sec^2 * (2.8125 sec)^2

s = 253.125 ft + 128 ft

s = 381.125 ft

Therefore, when it hits the ground, the net distance it has traveled is approximately 381.125 ft.

d. When it hits the ground, what is the total distance it has traveled?

To find the total distance traveled by the projectile, we can consider that the total distance up and down is the same.

Total distance = 2 * net distance traveled (since it goes up and down)

Total distance = 2 * 381.125 ft

Total distance = 762.25 ft

Therefore, when it hits the ground, the total distance traveled by the projectile is approximately 762.25 ft.