Question

A diver in Hawaii is jumping off a cliff 45m high, but she notices that there is an outcropping of rocks 7m out at the base. So, she must clear a horizontal distance of 7m during the dive in order to survive.

Assuming the diver jumps horizontally, what is his/her minimum push-off speed?

Answer 1

Answer:

u=2.33 m/s

Explanation:

Given that

Height ,h= 45 m

Horizontal distance ,x= 7 m

The initial speed in the vertical direction ,v= 0 m/s

The speed in the horizontal direction = u m/s

In the vertical direction :

$h=vt+\dfrac{1}{2}gt^2$

$45=0+\dfrac{1}{2}\times 10\times t^2$    ( take g= 10 m/s²)

t² = 9

t= 3 s

In the horizontal direction :

x = u t

7 = u x 3

$u=\dfrac{7}{3}\ m/s$  

u=2.33 m/s

Therefore the minimum push off speed will be 2.33 m/s.