a baseball leaves a bat with an initial speed of 37m/s at an angle of 53.1° find the position of the ball when t=2s( treat baseball as a projectile motion g = 9.8m/s^2)
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Answered by
1
Answer:
Explanation:
Assuming the baseball is in projectile,
Horizontal range R=u²*sin^theta/g
=37²*sin53.1/10(Taking g as 10 m/s²)
=1369*0.8/10
=1095.2/10
=109.52 m (Note: Value is different is g is 9.8)
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Solution
x=(v
0
cos53 ∘
)tx= 5
3
×37×2
x=44.4mby second eq
n
of motion,y=vsin(53
∘ )t− 2
g
t 2 y=(37)( 54 )(2)−(5)(2) 2 =39.2mv net (v)2 +( 53v 0
)
2
v
1
=u−gt [I
st
of motion]
v
1
=
5
uv
0
−(10)(2)=a.bm/s
So, v
net
=
(9.6)
2
+(22.2)
2
=24.4m/
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