Answer:
the magnitude of the ball's acceleration as it comes to rest on the foam is 817.5 m/s²
Explanation:
Given the data in the question;
initial velocity; u = 0 m/s
height; h = 2.5 m
we find the velocity of the ball just before it touches the foam.
using the equation of motion;
v² = u² + 2gh
we know that acceleration due gravity g = 9.81 m/s²
so we substitute
v² = ( 0 )² + ( 2 × 9.81 × 2.5 )
v² = 49.05
v = √49.05
v = 7.00357 m/s
Now as the ball touches the foam
final velocity v₀ = 0 m/s
compresses S = 3 cm = 0.03 m
so
v₀² = v² + 2as
we substitute
( 0 )² = 49.05 + 0.06a
0.06a = -49.05
a = -49.05 / 0.06
a = -817.5 m/s²
Therefore, the magnitude of the ball's acceleration as it comes to rest on the foam is 817.5 m/s²
