Question

From a cliff 37.6 m high. At the level of the sea, a rock sticks out a horizontal distance of 12.12 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

Answer:

4.3 m/sec

Explanation:

Here height of cliff = y = 37.6 m

Gravitational acceleration = g = 9.8 m/sec2

vi = 0 m/s

Let's find the time which the diver will take if jumps from there!

Using formula

y = vit+1/2gt2

==> 37.6= 0 + 0.5 ×9.8×$t^{2}$

==>$t^{2}$= $\frac{37.6}{4.9}$

==> t = 2.8 sec

In this time the diver has to cover a horizontal distance of 12.12 m

If x = 12.12 m is the horizontal distance to be covered then using

x= Vx × t

==> Vx = x/t

==> Vx= 12.12/2.8 = 4.3 m/s