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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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Answer:

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3 years ago

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Svetllana [295]

Answer:

The final position made with the vertical is 2.77 m.

Explanation:

Given;

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The final position made with the vertical (Yf) after 1.3 seconds is calculated as;

$Y_f = V_yt - \frac{1}{2}g t^2\\\\Y_f = (8.5 \times 1.3 ) - (\frac{1}{2} \times 9.8 \times 1.3^2)\\\\Y_f = 11.05 \ - \ 8.281\\\\Y_f = 2.77 \ m$

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Answer:

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