Question

Measure how high you can jump then calculate your initial velocity. Jump Height= .749 metersBe sure to show your workvf = vi + at▲X = vit + ½ at^2vf^2 = vi^2 +2ax▲X = ½ (vi + vf)t

Answer 1

Given information:

Height covered;

`x=0.749\text{ m}`

The third equation of motion is given as,

`v^2_f=v^2_i+2ax\ldots(1)`

Here, v_f is the final velocity, v_i is the initial velocity, a is the acceleration (during upward motion a=-g; g is the acceleration due to gravity) and x is the jump height.

At maximum height the final velocity v_f will be zero i.e.,

`v_f=0`

Therefore, from equation (1),

`\begin{gathered} 0^2=v^2_i-2gx \\ v^2_i=2gx \\ v_i=\sqrt[]{2gx} \end{gathered}`

Substitute 9.81 m/s² for g and 0.749 m for x,

`\begin{gathered} x=\sqrt[]{2*(9.81\text{ m/s}^2)*(0.749\text{ m})} \\ \approx3.83\text{ m/s} \end{gathered}`

Therefore, the initial velocity is 3.83 m/s.