Answer:
h' = 55.3 m
Explanation:
First, we analyze the horizontal motion of the projectile, to find the time taken by the arrow to reach the orange. Since, air friction is negligible, therefore, the motion shall be uniform:
s = vt
where,
s = horizontal distance between arrow and orange = 60 m
v = initial horizontal speed of the arrow = v₀ Cos θ
θ = launch angle = 30°
v₀ = launch speed = 35 m/s
Therefore,
60 m = (35 m/s)Cos 30° t
t = 60 m/30.31 m/s
t = 1.98 s
Now, we analyze the vertical motion to find the height if arrow at this time. Using second equation of motion:
h = Vi t + (1/2)gt²
where,
Vi = Vertical Component of initial Velocity = v₀ Sin θ = (35 m/s)Sin 30°
Vi = 17.5 m/s
Therefore,
h = (17.5 m/s)(1.98 s) + (1/2)(9.81 m/s²)(1.98 s)²
h = 34.6 m + 19.2 m
h = 53.8 m
since, the arrow initially had a height of y = 1.5 m. Therefore, its final height will be:
h' = h + y
h' = 53.8 m + 1.5 m
<u>h' = 55.3 m</u>
Answer:
The amplitude of the resultant wave will be 0.
Explanation:
Suppose the first wave has an amplitude of A. Its angle is given as wt.
The second way will also have the same amplitude as that of first.
After the reflection, a phase shift of π is added So the wave is given as
Adding the two waves give
So the amplitude of the resultant wave will be 0.
6.84m
Explanation:
According to this formula
<span>speed of wave=wavelength×frequency</span>
you have
<span>wavelength=<span>speedfrequency</span>=<span><span>342 m<span>s<span>−1</span></span></span><span>50<span>s<span>−1</span></span></span></span>=<span>6.84 m</span></span>
Answer:
240000 mph² or miles/hour²
Explanation:
<em>Use the formula</em>
<h3>acceleration = change in velocity ÷ time</h3>
change in velocity = 300 mph - 0 mph (final velocity - intial velocity)
time = 0.00125 hours
<em>Substitute the values into the formula:</em>
acceleration = 300 ÷ 0.00125 = 240000 mph²
