Question

A diver leaves the end of a 4.0 m high diving board and strikes the water 1.3s later, 3.0m beyond the end of the board. Considering the diver as a particle, determine: the magnitude of her initial velocity.

Answer 1

4.0 m/s

Step-by-step 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

`d=v_x t`

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,

`v_x = (d)/(t)=(3.0 m)/(1.3 s)=2.3 m/s`

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

`y(t) = h + v_y t - (1)/(2)gt^2`

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

`0=h+v_y t - (1)/(2)gt^2\\v_y = (0.5gt^2-h)/(t)=(0.5(9.8 m/s^2)(1.3 s)^2-4.0 m)/(1.3 s)=3.3 m/s`

So now we can find the magnitude of the initial velocity:

`v=√(v_x^2+v_y^2)=√((2.3 m/s)^2+(3.3 m/s)^2)=4.0 m/s`