Apole-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

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

The magnitude of his accelerations on the pad is 66.64 m/s²

Further explanation

Acceleration is rate of change of velocity.

a = acceleration ( m/s² )

v = final velocity ( m/s )

u = initial velocity ( m/s )

t = time taken ( s )

d = distance ( m )

Let us now tackle the problem!

Given:

height of pole vaulter above the pad = H = 4.2 - 0.8 = 3.4 m

distance of compression of pad = d = 50 cm = 0.5 m

Unknown:

a = ?

Solution:

Firstly , we calculate the velocity of pole-vaulter before hitting the pad.

Next , we can calculate the magnitude of his accelerations on the pad.

Learn moreVelocity of Runner : Kinetic Energy : Acceleration : The Speed of Car : Answer details

Grade: High School

Subject: Physics

Chapter: Kinematics

Keywords: Velocity , Driver , Car , Deceleration , Acceleration , Obstacle

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