Question

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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Answer 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)

Answer 2

HOPE IT WILL HELP YOU

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/