A) The velocity of the ball just before it strikes the floor : - 5.42 m/s
B) The velocity just before it strikes the floor : 5.33 m/s
C) Acceleration during contact with the floor : 1.34 * 10⁵ m/s²
D) Amount of compression during collision with floor : 1.09 * 10⁻⁴ m
Given data :
Height from which ball is dropped = 1.5 m
Rebounded height = 1.45 m
A) Velocity of ball just before it strikes the floor
v² = u² + 2as ----- ( 1 )
where : s = -1.5 m, a = -9.81 m/s², u = 0 m/s
insert values into equation ( 1 )
v = ±[tex]\sqrt{29.4 m^2/s^2}[/tex]
= - 5.42 m/s
B) Velocity just after the ball leaves the floor backup
v² = u² + 2as
where : s = 1.45m, a = - 9.8 m/s, v = 0 m/s , u = ?
insert values into equation above
u = [tex]\sqrt{28.42 m^2/s^2}[/tex]
= 5.33 m/s
C) Acceleration during contact with the floor
v = u + at ----- ( 3 )
where : v = 5.33 m/s , u = - 5.42 m/s, a = ? , t = 8 * 10⁻⁵ s
insert values into equation ( 3 )
a = ( 10.75 m/s ) / ( 8 * 10⁻⁵ s )
= 1.34 * 10⁵ m/s²
D) Amount of compression during collision with floor
v² = u² + 2as
where : v = -5.42 m/s , u = 0 m/s, a = 1.34 * 10⁵ m/s², s = ?
insert values into equation above
s = (29.38 m²/s² ) / ( 2.68 * 10⁵ m/s² )
= 1.09 * 10⁻⁴ m
Hence we can conclude that A) The velocity of the ball just before it strikes the floor : - 5.42 m/s
B) The velocity just before it strikes the floor : 5.33 m/s
C) Acceleration during contact with the floor : 1.34 * 10⁵ m/s²
D) Amount of compression during collision with floor : 1.09 * 10⁻⁴ m
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