A 0.30kg softball has a velocity of 15m/s at anangle of 35 degree below the horizontal just before making contactwith the bat. What is the magnitude o

York

York

Answered question

2020-11-09

A 0.30 kg softball has a haste of 15m/ s at anangle of 35 degree below the vertical just before making contactwith the club. What's the magnitude of the change in instigation ofthe ball while it's in contact with the club if the ball leaves thebat with a haste of( a) 20m/ s, haste over, and( b) 20m/ s, horizontally back toward the pitcher? 
a) π=0.3kg(15ms)cos(35)i 
b) π=0.320ms=6ms 
Ive

Answer & Explanation

Daphne Broadhurst

Daphne Broadhurst

Skilled2020-11-10Added 109 answers

The initial velocity of the pitcher
V1=15cos35i+15sin35j
=12.2872i+8.6036j
=14.9999ms
final velocity is given by
V2=20cos90i+20sin90j
=20j
=20
change in momentum is
=m(V2V1)=15.0001 kgm/s
b) IF the ball is sent back then
V2=20ms
change in momentum is
=m(V2V1)=0.3(2015)
= -105 kgm/s.

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