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givi [52]
3 years ago
7

90 points whats the equation for how fast something will be traveling before it hits the ground. How to do the equation PLEASE E

XPLAIN here is an example problem
A 0.05 kg-coin is dropped from the top of a skyscraper that is 100 meters high. How fast is the coin traveling just before it hits the ground?
use GPE=M*g*h
KE= 0.5*Mass *velocity squared

message me for questions
Physics
2 answers:
ArbitrLikvidat [17]3 years ago
7 0
⭐Hey

⭐Here we can use the formula for the body that is in free fall state. .

⭐that is

↪V = underoot 2*g*h

↪So the velocity will be ...

↪V = underoot 2 * 10 * 100

↪V = underoot 2000

↪V = 20 *underoot 5

↪V = 20 ( 2.23)

↪V = 44.6 m/s
qwelly [4]3 years ago
3 0

<em>You wrote the whole answer right there !</em>

The GPE it has when it's dropped IS exactly the KE it has when it hits the ground.  So you just write (KE at the bottom) = (GPE at the top).

Like this:   M · g · h  =  (1/2) · m · v²

Divide each side by M :

g · h  =  (1/2) · v²

Multiply each side by 2 :

2 · g · h  =  v²

Take the square root of each side:

<em>Speed = √(2 · g · h)</em>

What things do you know in this equation ?

-- 'g' is 9.81 m/s².  

-- 'h' is the height it dropped from.  

-- You know what '2' is.

-- You know how to do a square root on your calculator.

-- Pluggum all in, and you have the speed when it hits the ground.

Notice that the mass of the object went away !   It doesn't matter whether it's a coin or a car.  It only depends on gravity and the height it's dropped from.

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A baseball is hit at ground level. The ball reaches its maximum height above ground level 3.0 s after being hit. Then 2.5 s afte
vaieri [72.5K]

Answer:

Part a)

H = 44.1 m

Part b)

y = 13.48 m

Part c)

d = 8.86 m

Explanation:

Part a)

As we know that ball will reach at maximum height at

t = 3 s

now we will have

t = \frac{v sin\theta}{g}

now we have

3 = \frac{vsin\theta}{9.8}

v sin\theta = 29.4 m/s

Now maximum height above ground is given as

H = \frac{v^2sin^2\theta}{2g}

H = \frac{29.4^2}{2(9.8)}

H = 44.1 m

Part b)

Height of the fence is given as

y = (vsin\theta) t - \frac{1}{2}gt^2

y = (29.4)(5.5) - \frac{1}{2}(9.8)(5.5^2)

y = 13.48 m

Part c)

As we know that its horizontal distance moved by the ball in 5.5 s is given as

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97.5 = v_x (5.5)

v_x = 17.72 m/s

now total time of flight is given as

T = 3 + 3 = 6 s

so range is given as

R = v_x T

R = (17.72)(6)

R = 106.4 m

so the distance from the fence is given as

d = 106.4 - 97.5

d = 8.86 m

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