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Projectile questions knock youself out.?
1)(a) Prove that for a projectile fired from level ground at an angle θ above the horizontal, the ratio of the maximum height H to the range R is given by H/R = ¼ tan θ . (b) For what angle θ does H= R.
2)A movie stuntman is to run across and directly off a rooftop, to land on the roof of the next building 6.2 m away, 4.8m lower from the other building. Can he land there if he runs at 4.5m/s?
3) An archer tries to hit a target that is 20 m away from him. He can release the arrow at 25 m/s. Neglecting air resistance, estimate the angle at which the archer should aim to compensate for the fall of the arrow due to gravity
4)A golfer hits his tee shot along a flat fairway at 40o to the horizontal with an initial speed of 50 m/s. What is the range of the ball? What is the maximum height to which the ball rises?
5) A stone is thrown horizontally from the top of the building 440 m high. (a) Find its velocity after 3 seconds (b) How far is it from the ground at this time?
1)
½R = VcosΘ x T ---->(1)
Where T is time taken to reach max height .
(VsinΘ)² = 2gH . --> (2)
H = -½gT² + VsinΘxT --->(3)
VsinΘ = gT ----> (4)
Substitute (4) in (2)
(gT)² = 2gH
gT² = 2H -----> (5)
Substitute (5) in (3)
H = -½(2H) + VsinΘ x T
2H = VsinΘ x T
TanΘ = SinΘ/CosΘ
sinΘ = 2H/VT
cosΘ = ½R/VT
TanΘ = (2H/VT)/(½R/VT) = 4H/R
H/R = ¼TanΘ .
b)
When H = R
¼TanΘ = 1
TanΘ = 4
Θ = 75.96°
2)
First lets calculate the time it will take him to fall a distance of 4.8m Vertically .
Vertical Motion :-
s = ½at² + ut
-4.8 = ½(-9.8)t² + 0
t = 0.9897433186 s
Min Speed Required = Distance / Time Taken
= 6.2 / 0.9897433186 = 6.264250421m
So no , he wont make it .
3)
It means that vertical displacement should be equal to zero .
s = ½at² + ut
0 = ½(-9.8)t² + 25sinΘxt
20 = 25cosΘ x t
t = 20/(25cosΘ)
0 = ½(-9.8)(20/25cosΘ)² + 25sinΘ(20/25cosΘ)
0 = -392/125 sec²Θ + 20TanΘ
sec²Θ = Tan²Θ + 1
0 = -392/125 Tan²Θ + 20TanΘ - 392/125
TanΘ = 6.216693814
TanΘ = 0.1608572064
Θ = 80.86184113 ( Not appropriate )
Θ = 9.138158873 ANS 
4)
Horizontal Component Velocity = 50cos40
Vertical Component Velocity = 50sin40
R is range and H is max Height
At max height V=0
0 = 50sin40 - 9.8T
T = 3.279528621 s
H = ½(-9.8)( 3.279528621)² + 50sin40(3.279528621)
H = 52.70285996 m
R = 50cos40 x 2T
R = 50cos40 x 2(3.279528621) = 251.2264676 m
5)
I think there is an info for speed of projection is missing or you meant thrown vertically instead of horizontally . Please Check .
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