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Alex_Xolod [135]
4 years ago
12

How much time would it take for an

Physics
1 answer:
Elina [12.6K]4 years ago
4 0

36 and a half hours

Explanation:

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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
A diver jumps from a high platform and stays in the air for 1 second. How high was the platform?
ivolga24 [154]

Answer: Depends

Explanation:

Depends on how much the diver weighs.

6 0
2 years ago
An electromagnetic wave has a frequency of 5.0 x 1014 Hz. What is the
andrezito [222]

Answer:

c

Explanation:

wavelength = speed of light/ frequency

= (3x 10^8 m/s)/(5.0 x 10^14 Hz)

= 6.0 x 10^-7 m

5 0
3 years ago
A 90.0 kg person is being pulled away from a burning building as shown in the figure below.
Molodets [167]

Answer:

Try doing 90.0/2xT1 and that soud get u the answer of T2

Explanation:

I hope that helps!

6 0
3 years ago
Gulls are often observed dropping clams and other shellfish from a height to the rocks below, as a means of opening the shells.
kumpel [21]

Answer:

v = 17.71 m / s

Explanation:

We can work this exercise with the kinematics equations. In general the body is released so that its initial velocity is zero, the acceleration of the acceleration of gravity

                v² = v₀² - 2 g (y -y₀)

                v² = 0 - 2g (y -y₀)

when it hits the stone the height is zero and part of the height of the seagull I

              v² = 2g y₀

              v = Ra (2g i)

let's calculate

              v =√ (2 9.8 16)

              v = 17.71 m / s

8 0
3 years ago
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