Question

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?

Answer 1

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!