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

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:

$t_{v} = \sqrt{\frac{2h}{g}}$

Where:

h: is the height = 10m

g: si the gravity = 9.81 m/s²

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

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!