Answer:
3.6ft
Explanation:
Using= 2*π*sqrt(L/32)
To solve for L, first move 2*n over:
T/(2*π) = sqrt(L/32)
Next,eliminate the square root by squaring both sides
(T/(2*π))2 = L/32
or
T2/(4π2) = L/32
Lastly, multiply both sides by 32 to yield:
32T2/(4π2) = L
and simplify:
8T²/π²= L
Hence, L(T) = 8T²/π²
But T = 2.1
Pi= 3.14
8(2.1)²/3.14²
35.28/9.85
= 3.6feet
Assume no air resistance, and g = 9.8 m/s².
Let
x = angle that the initial velocity makes with the horizontal.
u = 30 cos(x), horizontal velocity
v = 30 sin(x), vertical launch velocity
The horizontal distance traveled is 55 m, therefore the time of flight is
t = 55/[30 cos(x)] = 1.8333 sec(x) s
With regard to the vertical velocity, and the time of flight,obtain
[30 sin(x)]*(1.8333 sec(x)) + (1/2)*(-9.8)*(1.8333 sec(x))² = 0
55 tan(x) - 16.469 sec²x = 0
55 tan(x) - 16.469[1 + tan²x] = 0
16.469 tan²x - 55 tan(x) + 16.469 = 0
tan²x - 3.3396 tan(x) + 1 = 0
Solve with the quadratic formula.
tan(x) = 0.5[3.3396 +/- √(7.153)] = 3.007 or 0.3326
Therefore
x = 71.6° or x = 18.4°
The time of flight is
t = 1.8333 sec(x) = 5.8096 s or 1.932 s
The initial vertical velocity is
v = 30 sin(x) = 28.467 m/s or 9.468 m/s
The horizontal velocity is
u = 30 cos(x) = 9.467 m/s or 28.469 m/s
If t = 5.8096 s,
u*t = 9.467*5.8096 = 55 m (Correct)
or
u*t = 28.469*15.8096 = 165.4 m (Incorrect)
Therefore, reject x = 18.4°. The correct solution is
t = 5.8096 s
x = 71.6°
u = 9.467 m/s
v = 28.467 m/s
The height from which the ball was thrown is
h = 28.467*5.8096 - 0.5*9.8*5.8096² = -110.4 m
The ball was thrown from a height of 110.4 m
Answer: h = 110.4 m
According to my examination I have confirmed that I do NOT repeat do NOT know this .
Answer:
hope this helped you
please mark as the brainliest (ㆁωㆁ)
Light travels in waves AND in bundles called "photons".
It's hard to imagine something that's a wave and also a bundle.
But it turns out that light behaves like both waves and bundles.
If you design an experiment to detect waves, then it responds to light.
And if you design an experiment to detect 'bundles' or particles, then
that one also responds to light.
