Answer:
4.0 m/s
Explanation:
The motion of the diver is the motion of a projectile: so we need to find the horizontal and the vertical component of the initial velocity.
Let's consider the horizontal motion first. This motion occurs with constant speed, so the distance covered in a time t is
[tex]d=v_x t[/tex]
where here we have
d = 3.0 m is the horizontal distance covered
vx is the horizontal velocity
t = 1.3 s is the duration of the fall
Solving for vx,
[tex]v_x = \frac{d}{t}=\frac{3.0 m}{1.3 s}=2.3 m/s[/tex]
Now let's consider the vertical motion: this is an accelerated motion with constant acceleration g=9.8 m/s^2 towards the ground. The vertical position at time t is given by
[tex]y(t) = h + v_y t - \frac{1}{2}gt^2[/tex]
where
h = 4.0 m is the initial height
vy is the initial vertical velocity
We know that at t = 1.3 s, the vertical position is zero: y = 0. Substituting these numbers, we can find vy
[tex]0=h+v_y t - \frac{1}{2}gt^2\\v_y = \frac{0.5gt^2-h}{t}=\frac{0.5(9.8 m/s^2)(1.3 s)^2-4.0 m}{1.3 s}=3.3 m/s[/tex]
So now we can find the magnitude of the initial velocity:
[tex]v=\sqrt{v_x^2+v_y^2}=\sqrt{(2.3 m/s)^2+(3.3 m/s)^2}=4.0 m/s[/tex]
