Question

A 51.9 kg pole vaulter running at 10.2 m/s vaults over the bar. The acceleration of gravity is 9.81 m/s 2 . If the vaulter’s horizontal component of velocity over the bar is 1.1 m/s and air resistance is disregarded, how high is the jump? Answer in units of m.

Answer 1

Answer:

h =  5.25 m

Explanation:

given,

mass of the vaulter, m = 51.9 Kg

speed of the vaulter, v = 10.2 m/s

acceleration due to gravity, a =9.81 m/s²

Speed when he is in air = 1.1 m/s

height of the jump = ?

to solve this question we will use conservation of energy

 $KE_i + PE_i = KE_f + PE_f$

initial potential energy is equal to zero

 $\dfrac{1}{2}mv_i^2 + 0 = \dfrac{1}{2}mv_f^2+ m g h$

 $h = \dfrac{v_i^2-v_f^2}{2g}$

 $h = \dfrac{10.2^2-1.1^2}{2\times 9.8}$

     h =  5.25 m

Vaulter jump to the height of 5.25 m.