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3 years ago

A box falls out of a stationary helicopter hovering 135 m above the ground. How long will it take to hit the ground?

1 answer:

4 0

We have the equation of motion $s = ut + \frac{1}{2} at^2$ , where s is the displacement, a is the acceleration, u is the initial velocity and t is the time taken.

Here displacement = 135 m, Initial velocity = 0 m/s, acceleration = 9.81 $m/s^2$

Substituting

$135 = 0*t+\frac{1}{2} *9.8*t^2\\ \\ 4.9t^2 =135\\ \\ t =5.25 seconds$

A box falls out of a stationary helicopter hovering 135 m above the ground will take 5.25 seconds to reach the ground.

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A 2.0 kg puck is at rest on a level table. It is pushed straight north with a constant force of 5N for 1.50 s and then let go. H

lukranit [14]

Answer:

d = 6.32 m

Explanation:

Given that,

The mass of a puck, m = 2 kg

It is pushed straight north with a constant force of 5N for 1.50 s and then let go.

We need to find the distance covered by the puck when move from rest in 2.25 s.

We know that,

F = ma

$a=\dfrac{F}{m}\\\\a=\dfrac{5}{2}\\\\a=2.5\ m/s^2$

Let d is the distance moved in 2.25 s. Using second equation of motion,

$d=ut+\dfrac{1}{2}at^2\\\\d=0+\dfrac{1}{2}\times 2.5\times (2.25)^2\\\\d=6.32\ m$

So, it will move 6.32 m from rest in 2.25 seconds.

4 0

3 years ago

PLEASE HELP

Sergeu [11.5K]

The vertical component of the initial velocity is $v_0_y = \frac{y}{t} + \frac{1}{2} gt$

The horizontal component of the initial velocity is $v_0_x = \frac{x}{t}$

The horizontal displacement when the object reaches maximum height is $X = \frac{xy}{gt^2} + \frac{x}{2}$

The given parameters;

the horizontal displacement of the object, = x

the vertical displacement of the object, = y

acceleration due to gravity, = g

time of motion, = t

The vertical component of the initial velocity is given as;

$y = v_0_yt - \frac{1}{2} gt^2\\\\v_0_yt = y + \frac{1}{2} gt^2\\\\v_0_y = \frac{y}{t} + \frac{1}{2} gt$

The horizontal component of the initial velocity is calculated as;

$x = v_0_xt\\\\v_0_x = \frac{x}{t}$

The time to reach to the maximum height is calculated as;

$T = \frac{v_f_y -v_0_y}{-g} \\\\T = \frac{-v_0_y}{-g} \\\\T = \frac{v_0_y}{g} \\\\T = \frac{1}{g} (v_0_y)\\\\T = \frac{1}{g} (\frac{y}{t} + \frac{1}{2} gt)\\\\T = \frac{y}{gt} + \frac{1}{2} t$

The horizontal displacement when the object reaches maximum height is calculated as;

$X= v_0_x \times T\\\\X= \frac{x}{t} \times (\frac{y}{gt} + \frac{1}{2} t)\\\\X = \frac{xy}{gt^2} + \frac{x}{2}$

Learn more here: brainly.com/question/20689870

8 0

3 years ago

Consider Newton's Law of Universal Gravitation: FG= G (m1 m2)/d2 .

bixtya [17]

Answer:

Mass is directly proportional to the Force of Gravity. If Mass increases, then the Force of Gravity increases; however, Distance is indirectly (or inversely) proportional to the Force of Gravity. If Distance increases, then the Force of Gravity decreases.

Explanation:

The formula for the force of gravity between two objects is

$F=G\frac{m_1 m_2}{d^2}$

where

G is the gravitational constant

m1, m2 are the masses of the two objects

d is the separation between the two objects

We notice the following:

- F is directly proportional to the masses, $F\propto m_1, m_2$ . This means that if one of the masses increases, then the force between them, F, increases in a proportional way

- F is inversely proportional to the square of the distance, $F\propto \frac{1}{d^2}$ . This means that if the distance between the two objects is increased, the force between them will decrease, and vice-versa.