A bowler throws a ball horizontally along east direction
withspeedof 144km/hr. Thebatsman hitstheballsuchthat it
deviates from its initial direction ofmotion by 74° north of east
1
direction, without changing its speed. Ifmass ofthe ball is j kg
and time ofcontact between bat and ball is 0.02
Answers
Answered by
2
Answer:
y
Explanation:
if it is 144km mph then do this
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Answered by
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The force is acting 800 N, 53° west of north.
Explanation:
- Speed of bowler = 144 Km/hr
- Direction = 74° north of east
- Mass of ball = 1/3 Kg
- Time of contact = 0.02 seconds
Solution:
F = dp / dt
v1 = 40 m/s
v2 = 40 m/s
| v2 - v1 | = √ 40^2 + 40^2 - 2 (40)^2 cos 74°
| v2 - v1 | = 40 √ 2 - 2 cos 74°
cos 2 x 37 = 40 √ ( 1 - cos 74° )
1 - sin^2 θ = 40 √ 2 ( 1 - cos 74° )
1 - sin^2 θ = 40 √ [ 1- ( 1 - 2 sin^2 37° )
| v2 - v1 | = 40 x 2 x sin 37°
| v2 - v1 | = 40 x 2 x 3/5
| v2 - v1 | = 48 m /s
ΔP = 48 x 1/3 = 16
F = ΔP / t = 16 / 0.02 = 800 N
The force is acting 800 N, 53° west of north.
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