Physics, asked by shivasinghmohan629, 20 days ago

I am having problem for one question ​

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Answers

Answered by dagaridevi1153
0

Answer:

It looks like there aren't many great matches for you

Explanation:

hope it's helpful

Answered by syedsufiyan5
0

Answer:

u = 29.4 m s − initial speed of the ball;

h = 4.9m − maximum height that ball riches

L − range of a ball;

Equation of the motion for body, thrown at angle α:

(L - maximum horizontal range of this body)

(t − time of the flight)

ux = u cos α; uy = u sin α;

x: L = ut cos α (1)

y: 0 = ut sin α −gt²/2

u sin α =gt/2

t = 2u sin α - /g (2)

(2) in (1):

L = u -2u sin α/g

cos α = 2u² sin α cos α/g (3)

Maximum height: the time taken to reach the maximum height is equal to half of the time of

flight:

t1 =

t

2

=

u sin α

g

y: (half of the flight): h1 = ut1 sin α −

gt1

2

2

h = ut1 sin α −

gt1

2

2

h = u

V sin α

g

sin α −

g (

u sin α

g

)

2

2

=

u

2

sin2 α

2g (4)

from (4)

sin2 α =

2gh

u

2

sin α = √

2gh

u

2

(5)

Pythagorean Identity

sin2 α + cos2 α = 1

cos2 α = 1 − sin2 α (6)

(5)in(6):

cos2 α = 1 −

2gh

u

2

cos α = √1 −

2gh

u

2

(7)

(5)and(7)in(3):

L =

2u

2

sin α cos α

g

=

2u

2√(

2gh

u

2

) (1 −

2gh

u

2

)

g

=

2√(2gh)(u

2 − 2gh)

g

=

=

2√(2 ∙ 9.8

m

s

2

∙ 4.9m) ((29.4

m

s

)

2

− 2 ∙ 9.8

m

s

2

∙ 4.9m)

9.8

m s ²

= 55 m

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