A 72.1 kg high jumper leaves the ground with a vertical velocity of 9.3 m/s. How high can he jump? The acceleration of gravity is 9.8 m/s 2 . Answer in units of m

4.408 m.

Explanation:

Given:

m = 72.1 kg

vf = 9.3 m/s

vi = 0m/s

a = 9.81 m/s^2

Equations of motion:

i. vf = vi + a*t

ii. S = vi*t + 1/2*(a*(t^2))

iii. vf^2 = vi^2 + 2a*S

Using iii. equation of motion;

(9.3)^2 = 0 + 2 * (9.8) * S

S = 86.49/19.62

= 4.408 m.

0.93xc=0.93x3x10^8m/s

2.54cm is 1 inch

360seconds i 1 hr