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?

Question

asked Aug 15, 2021 112k views

1 Answer

Nakeuh · Answer 1 · 2021-08-19T12:45:22+0000

Answer:

The distance reached is 14.3 m.

Step-by-step explanation:

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

$t_(v) = \sqrt{(2h)/(g)}$

Where:

h: is the height = 10m

g: si the gravity = 9.81 m/s²

$t_(v) = \sqrt{(2h)/(g)} = \sqrt{(2*10 m)/(9.81 m/s^(2))} = 1.43 s$

Now, we can find the distance reached:

$x = V_{0_(x)}*t_(v)$

Where:

$V_{0_(x)}$ is the initial horizontal velocity = 10 m/s

$x = V_{0_(x)}*t_(v) = 10 m/s*1.43 s = 14.3 m$

Therefore, the distance reached is 14.3 m.

I hope it helps you!