Question

A 64.0 kg pole vaulter running at 10.2 m/s vaults over the bar. If the vaulter's horizontal component of velocity over the bar is 1.0 m/s and air resistance is disregarded, how high was the jump?

Answer 1

Answer:

h = 5.05 m

Explanation:

given,

mass of pole vaulter, m = 64 Kg

speed of the vaulter,V = 10.2 m/s

horizontal component of velocity in air, v = 1 m/s

height of the jump,h = ?

using energy conservation

     $E_i = E_f$

$\dfrac{1}{2}mv_i^2 + m g h_i= \dfrac{1}{2}mv_f^2+m g h_f$

initial height of the vaulter is equal to zero.

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

$h =\dfrac{v_i^2-v_f^2}{2g}$

$h =\dfrac{10^2-1^2}{2\times 9.8}$

 h = 5.05 m

height of the jump is equal to 5.05 m.