Question

If the diver in item 2 leaves the board with an initial upward speed of 2.00 m/s, find the diver's speed when striking the water.

Answer 1

Answer:

$v_f=9.0774m/s$

Step-by-step explanation:

Let's assume

initial velocity =vi

final velocity=vf

initial height =hi

final height =hf

acceleration due to gravity =g

we can use formula

$\frac{1}{2}v_i^2+gh_i= \frac{1}{2}v_f^2+gh_f$

we are given

$v_i=2m/s$

$h_i=2\times 2=4$

$h_f=0$

$g=9.8m/s^2$

now, we can plug values

$\frac{1}{2}(2)^2+(9.8)(4)= \frac{1}{2}v_f^2+(9.8)(0)$

now, we can solve for vf

$\frac{1}{2}(2)^2+(9.8)(4)= \frac{1}{2}v_f^2$

we get

$v_f=9.0774m/s$