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Answer:
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Explanation:
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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