A 54.0 kg pole vaulter running at 8.3 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

Hi Charlotte A.!

An efficient solution is to consider the conservation of energy:

E_o + W_nc = E_f

Because we are neglecting air resistance, there are not other obvious forms of non-conservative work that can rob our system of energy. As a result, we can drop that term from our equation:

E_o = E_f

We must consider which energies we have at each point. For instance, our pole vaulter initially begins from the ground, so we can take the height equal to 0m. Consequently, we can also say there is no initial gravitational potential energy.

Furthermore, we are interested in knowing what happens at the maximum height. In such a case, we will have a gravitational potential energy. However, because our pole vaulter is still moving horizontally at this point, we can also deduce that there is a kinetic energy value as well.

So, we have:

KE_o = KE_f + Ug_f

(1/2)mv_o² = (1/2)mv_f² + mgh_max

Note that the mass of our pole vaulter cancels because it appears in every term:

(1/2)v_o² = (1/2)v_f² + gh_max

h_max = [(1/2)v_o² - (1/2)v_f²]/g = (v_o² - v_f²⁾/2g

Here, you might be wondering how do we know the final velocity? All we are given is the horizontal component. But let us remember that the vertical component of velocity is equal to 0 m*s^-1 at max height. Which means the horizontal component is the only velocity.

<=> [(8.3 m*s^-1)² - (1.0 m*s^-1)²]/(2*9.8 m*s^-2)

h_max = 3.5 m

Hope this helps!

Cheers