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4 years ago

An object is dropped from a platform 100 feet high. Ignoring wind resistance, what will its speed be when it reaches the ground?

2 answers:

7 0

if we assume the origin is at the dropping point and the object is merely dropped and not thrown up or down then y0 = 0 and v0 = 0. The equation reduces to 

y = 0 + 0t + ½gt² 
y = ½gt² 

t = √(2y/g) 

in the ft - lb - s system 

y = -100 ft 
g = -32.2 ft / s² 

t = √(2y/g) 
t = √(2(-100) / (-32.2)) 
t = 2.5 s

Irina-Kira [14]4 years ago

4 0

the answer to the question is 2.5

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I would like help with this physics problem

Darina [25.2K]

(a) This is a freefall problem in disguise - when the ball returns to its original position, it will be going at the same speed but in the opposite direction. So the ball's final velocity is the negative of its initial velocity.

Recall that

$v_f=v_i+at$

We have $v_f=-v_i$ , so that

$-2v_i=at\implies-2\left(8\,\dfrac{\mathrm m}{\mathrm s}\right)=\left(-2\,\dfrac{\mathrm m}{\mathrm s^2}\right)t\implies t=8\,\mathrm s$

(b) The speed of the ball at the start and at the end of the roll are the same 8 m/s, so the average speed is also 8 m/s.

(d) The position of the ball $x_f$ at time $t$ is given by

$x_f=x_i+v_it+\dfrac12at^2$

Take the starting position to be the origin, $x_i=0$ . Then after 6 seconds,

$x_f=\left(8\,\dfrac{\mathrm m}{\mathrm s}\right)(6\,\mathrm s)+\dfrac12\left(-2\,\dfrac{\mathrm m}{\mathrm s^2}\right)(6\,\mathrm s)^2=42\,\mathrm m$

so the ball is 42 m away from where it started.

We're not asked to say in which direction it's moving at this point, but just out of curiosity we can determine that too:

$x_f-x_i=\dfrac{v_i+v_f}2t\implies42\,\mathrm m=\dfrac{8\,\frac{\mathrm m}{\mathrm s}+v_f}2(6\,\mathrm s)\implies v_f=6\,\dfrac{\mathrm m}{\mathrm s}$

Since the velocity is positive, the ball is still moving up the incline.

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Which of the following does not affect electrical resistance?

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3 years ago

1. A small package is dropped from the Golden Gate Bridge. What is the velocity of the package

Sidana [21]

The velocity of the package after it has fallen for 3.0 s is 29.4 m/s

From the question,

A small package is dropped from the Golden Gate Bridge.

This means the initial velocity of the package is 0 m/s.

We are to calculate the velocity of the package after it has fallen for 3.0 s.

From one of the equations of kinematics for objects falling freely,

We have that,

v = u + gt

Where

v is the final velocity

u is the initial velocity

g is the acceleration due to gravity

and t is time

To calculate the velocity of the package after it has fallen for 3.0 s

That means, we will determine the value of v, at time t = 3.0 s

The parameters are

u = 0 m/s

g = 9.8 m/s²

t = 3.0 s

Putting these values into the equation

v = u + gt

We get

v = 0 + (9.8×3.0)

v = 0 + 29.4

v = 29.4 m/s

Hence, the velocity of the package after it has fallen for 3.0 s is 29.4 m/s

Learn more here: brainly.com/question/13327816

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