MARKING BRAINLIEST AND I NEED WORK SHOWN ! i’m a junior in high school Free Fall Practice Problems

1. A baseball is popped straight up in the air at a velocity of 42.0 m/s.

What time does it spend in the air before reaching the same

elevation from which it was popped up?

2. A golf ball is dropped from a cliff and strikes the ground with a

downward velocity of 34.0 m/s. How high is the cliff?

3. An air rocket is launched straight upwards at 27.0 m/s. What is its

velocity at time t=4.80 s?

4. A giant lizard jumps straight upwards from the ground at 4.30 m/s.

To what height does the lizard rise before going back down again?

5. A person falls from a bridge that is 18.0 m above the water. What

is the velocity of impact with the water?

(1) t = 4.29 s

(2) h = 58.98 m

(3) v = 20.04 m/s (downward)

(4) h = 0.94 m

(5) v = 18.78 m/s

Explanation:

Assumption:

(1)

initial velocity of the baseball, u = 42.0 m/s

final velocity of the ball, v = 0

time of flight is given by;

v = u -gt

gt = u

t = u / g

t = 42 /9.8

t = 4.29 s

(2)

Given;

initial velocity of the golf, u = 0

final velocity of the golf, v = 34 m/s

the height of the cliff is given by;

v² = u² + 2gh

v² = 0 + 2gh

h = v²/2g

h = (34)²/(2x9.8)

h = 58.98 m

(3)

Given;

initial velocity of the rocket, u = 27 m/s

time of motion, t = 4.8s

final velocity, v = ?

v = u + gt

v = 27 + (-9.8 x 4.8)

v = 27 - 47.04

v = -20.04 m/s (the negative sign indicates downward motion of the rocket after 4.8 s)

(4)

Given;

initial velocity of the lizard, u = 4.3 m/s

final velocity of the lizard at maximum height, v = 0

the height rose by the lizard is given by;

v² = u² + 2(-g)h

0 = u² - 2gh

2gh = u²

h = u² / 2g

h = (4.3)²/(2 x 9.8)

h = 0.94 m

(5)

Given;

initial velocity of the person on the bridge, u = 0

height of the bridge, h = 18 m

final velocity of the person is the velocity at impact, v = ?

v² = u² + 2gh

v² = 0 + 2gh

v² = 2gh

v = √2gh

v = √(2 x 9.8 x 18)

v = 18.78 m/s

gets less

when two balls of the different masses are dropped from equal height, they reach the ground at the same time .this is because they experience the same acceleration.

the acceleration is independent of mass.

thus when they are dropped they will always maintain the same velocity and travel the same distance (neglect air resistance of course) at the same time.