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3 years ago

Jason throws a ball horizontally off a 300 meter cliff at 15.0 m/s. How far will it travel before it hits the

Physics

2 answers:

jasenka [17]3 years ago

8 0

Answer:

117,30m

Explanation:

I think the situation here is a horizontal projection to the ground. So in order to find the distance the formula = Ut, where U is the initial speed and t is the Time of flight. To get the time of flight in this case =√2h/g where h is the height and g is gravity. so to get the time = √2×300÷9.81 =7.821 .so range =ut which is equal to the time multiplied by 15m/s =117.30m

Butoxors [25]3 years ago

5 0

Hi there!

To find the total distance traveled, we must find the time taken to fall down.

Ignore the horizontal velocity when dealing with the vertical direction. We can use the simplified equation:

$t = \sqrt{\frac{2h}{g}{}}$

Plug in the given values:

$t = \sqrt{\frac{2(300)}{9.81}} = 7.821 sec$

Use the equation:

Δd = vt

Plug in the time and horizontal velocity:

Δd = (15)(7.821) = 117.30m

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The mass of the Sun is 2 × 1030 kg, and the distance between Neptune and the Sun is 30 AU. What is the orbital period of Neptune

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Let's convert the radius of the orbit (the distance between the Sun and Neptune) from AU to meters. We know that 1 AU corresponds to 150 million km, so
$1 AU = 1.5 \cdot 10^{11} m$
so the radius of the orbit is
$r=30 AU = 30 \cdot 1.5 \cdot 10^{11} m=4.5 \cdot 10^{12} m$

And if we re-arrange the equation (1), we can find the orbital period of Neptune:
$T=\sqrt{ \frac{4 \pi^2}{GM} r^3} = \sqrt{ \frac{4 \pi^2}{(6.67 \cdot 10^{-11} m^3 kg^{-1} s^{-2} )(2\cdot 10^{30} kg)}(4.5 \cdot 10^{12} m)^3 }= 5.2 \cdot 10^9 s$

We can convert this value into years, to have a more meaningful number. To do that we must divide by 60 (number of seconds in 1 minute) by 60 (number of minutes in 1 hour) by 24 (number of hours in 1 day) by 365 (number of days in 1 year), and we get
$T=5.2 \cdot 10^9 s /(60 \cdot 60 \cdot 24 \cdot 365)=165 years$

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