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
2 answers:
Refer to the diagram below.
Ignoring air resistance, use gravitational acceleration g = 9.8 m/s².
The pole vaulter drops with an initial vertical speed u = 0.
At impact with the pad, velocity v satisfies:
v² = 2 × (9.8 m/s²) × (4.2 m) = 82.32 (m/s)²
v = 9.037 m/s
As the pad compresses by 0.5 m to bring the vaulter to rest,
let the average acceleration (deceleration) be a m/s². Then:
0 = (9.037 m/s)² + 2 × a × 0.5 m
Solving for a gives:
a = - 82.32 / (2 × 0.5) = -82 m/s²
Thus, the deceleration magnitude is 82 m/s².
Ignoring air resistance, use gravitational acceleration g = 9.8 m/s².
The pole vaulter drops with an initial vertical speed u = 0.
At impact with the pad, velocity v satisfies:
v² = 2 × (9.8 m/s²) × (4.2 m) = 82.32 (m/s)²
v = 9.037 m/s
As the pad compresses by 0.5 m to bring the vaulter to rest,
let the average acceleration (deceleration) be a m/s². Then:
0 = (9.037 m/s)² + 2 × a × 0.5 m
Solving for a gives:
a = - 82.32 / (2 × 0.5) = -82 m/s²
Thus, the deceleration magnitude is 82 m/s².
The magnitude of the pole-vaulter's acceleration on the pad is 66.64 m/s².
Explanation
Acceleration represents how quickly velocity changes.
a = acceleration (m/s²)
v = final speed (m/s)
u = initial speed (m/s)
t = time interval (s)
d = displacement (m)
Let's solve the problem step by step!
Given:
Height above pad: H = 4.2 m - 0.8 m = 3.4 m
Pad compression distance: d = 0.5 m
Find:
Acceleration, a
Solution:
First, calculate velocity just before hitting the pad.
Next, find the magnitude of acceleration decelerating the vaulter within the compressed pad.
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Response:
The new resistance is half of the original resistance.
Explanation:
Resistance in a wire is represented by:
= resistivity of the material
L and A are the physical dimensions
If a wire is exchanged for one where all linear dimensions are doubled, i.e. l' = 2l and r' = 2r
The updated resistance of the wire can be calculated as follows:
The new resistance equals half of the original resistance. Thus, this provides the solution needed.