Question

Cliff divers at Acapulco jump into the sea macias (fjm793) – Homework 3, 2d motion 19-20 – dowd – (WoffordWPHY11920 2) 4 from a cliff 36.2 m high. At the level of the sea, a rock sticks out a horizontal distance of 11.98 m. The acceleration of gravity is 9.8 m/s 2 . With what minimum horizontal velocity must the cliff divers leave the top of the cliff if they are to miss the rock? Answer in units of m/s.

Answer 1

v = 7.67 m/s

Step-by-step explanation:

Given data:

horizontal distance 11.98 m

Acceleration due to gravity 9.8 m/s^2

Assuming initial velocity is zero

we know that

`h = (gt^2)/(2)`

solving for t

we have

`t = \sqrt{(2h)/(g)}`

substituing all value for time t

`t = \sqrt{(2* 11.98)/(9.8)}`

t = 1.56 s

we know that speed is given as

`v = (d)/(t)`

`v =(11.98)/(1.56)`

v = 7.67 m/s