A pole-vaulter is nearly motionless as he clears the bar, set 4.2 m above the ground. he then falls onto a thick pad. the top of the pad is 80 cm above the ground, and it compresses by 50 cm as he comes to rest. what is the magnitude of his accelerations he comes to rest on the pad

Refer to the diagram shown below.

Neglect wind resistance, and use g = 9.8 m/s².

The pole vaulter falls with an initial vertical velocity of u = 0.

If the velocity upon hitting the pad is v, then

v² = 2 * (9.8 m/s²) * (4.2 m) = 82.32 (m/s) ²

v = 9.037 m/s

The pole vaulter comes to res after the pad compresses by 50 cm (or 0.5 m).

If the average acceleration (actually deceleration) is (a m/s²), then

0 = (9.037 m/s) ² + 2 * (a m/s²) * (0.5 m)

a = - 82.32 / (2*0.5) = - 82 m/s²

Answer: - 82 m/s² (or a deceleration of 82 m/s²)