The observation deck of a skyscraper is 420 m above
1 answer:
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Answer:
y = x tan θ - (g / 2v₀² cos² θ) x²
Explanation:
An equation is called a general formula that relates the position on the x-axis and the height on the body's axis.
Let's write the position on each axis
X axis. No acceleration
x = v₀ₓ t
Y Axis. There is the acceleration of gravity
y = t - ½ g t²
Let's clear the time in the first equation and substitute in the second
y = v₀ sin θ (x / v₀ₓ) - ½ g (x / v₀ₓ)²
y = v₀ sin θ / v₀ cos θ x - ½ g x² / v₀² cos² θ
y = x tan θ - (g / 2v₀² cos² θ) x²
This is the trajectory equation in projectile launching
Answer
given,
Time period= T = 1.5 s
If it's moving through equilibrium point at t₀= 0 with v = 1.0 m/s
v_max=1.00 m/s
we know,
v_ max=A ω
v = A sin (ωt)
-0.50= -1.00 sin (ωt)
sin (ωt) = 0.5
t = 0.125 s
we have time period T=1.5 it is the time to complete one oscillation
means from eq to right,then left,then eq,then left,then from right to eq
time taken for left = t/4 = 0.125/4 = 0.375 s
smallest value of time
=0.375 + 0.125
= 0.50 sec
Answer:371.564 mi
Explanation:
Given
Airplane flies northwest for 250 mi and then travels west 150 mi
That is first it travels 250cos45 in - ve x direction and simultaneously 250sin45 in y direction
after that it travels 150 mi in -ve x direction
So its position vector is given by
so magnitude of displacement is
|r|=371.564 mi