A cliff jumper jumps off a cliff with an initial horizontal velocity of 10 m/s. The cliff is 10 meters high. How far from the base of the cliff does the diamond fall?

Respuesta :

Answer:

The distance reached is 14.3 m.

Explanation:

To find the distance reached by the diamond first we need to find the flight time:

[tex] t_{v} = \sqrt{\frac{2h}{g}} [/tex]

Where:

h: is the height = 10m

g: si the gravity = 9.81 m/s²

[tex] t_{v} = \sqrt{\frac{2h}{g}} = \sqrt{\frac{2*10 m}{9.81 m/s^{2}}} = 1.43 s [/tex]

Now, we can find the distance reached:

[tex] x = V_{0_{x}}*t_{v} [/tex]

Where:

[tex]V_{0_{x}}[/tex] is the initial horizontal velocity = 10 m/s

[tex] x = V_{0_{x}}*t_{v} = 10 m/s*1.43 s = 14.3 m [/tex]

Therefore, the distance reached is 14.3 m.

I hope it helps you!