A) Using:
2as = v² - u², where v will be 0 at max height
s = -(160)² / 2 x -32.174
s = 397.8 ft
b) Using:
s = ut + 1/2 at²
256 = 160t - 16.1t²
solving for t,
t = 2.0, t = 7.9
Now, v = u + at, for both times:
v(2) = 160 - 32.174(2)
v(2) = 95.7 ft/sec on the way up
v(7.9) = 160 - 32.174(7.9)
v(7.9) = -94.3 ft/sec; 94.3 ft/sec on the way down
c) -32.174 ft/s², which is the acceleration due to gravity.
d) s = 0
0 = 160t - 1/2 x 32.174t²
t = 9.94 seconds
Answer:
X = 2146.05 m
Explanation:
We need to understand first what is the value we need to calculate here. In this case, we want to know how far from the starting point the package should be released. This is the distance.
We also know that the plane is flying a certain height with an specific speed. And the distance we need to calculate is the distance in X with the following expression:
X = Vt (1)
However we do not know the time that this distance is covered. This time can be determined because we know the height of the plain. This time is referred to the time of flight. And the time of flight can be calculated with the following expression:
t = √2h/g (2)
Where g is gravity acceleration which is 9.8 m/s². Replacing the data into the expression we have:
t = √(2*2500)/9.8
t = 22.59 s
Now replacing into (1) we have:
X = 95 * 22.59
<h2>
X = 2146.05 m</h2>
This is the distance where the package should be released.
Hope this helps
Answer:
the lowest possible frequency of the emitted tone is 404.79 Hz
Explanation:
Given the data in the question;
S₁ ← 5.50 m → L
↑
2.20 m
↓
S₂
We know that, the condition for destructive interference is;
Δr = ( 2m + 
) × λ
where m = 0, 1, 2, 3 .......
Path difference between the two sound waves from the two speakers is;
Δr = √( 5.50² + 2.20² ) - 5.50
Δr = 5.92368 - 5.50
Δr = 0.42368 m
v = f × λ
f = ( 2m + )v / Δr
m = 0, 1, 2, 3, ....
Now, for the lowest possible frequency, let m be 0
so
f = ( 0 + )v / Δr
f = (v) / Δr
we know that speed of sound in air v = 343 m/s
so we substitute
f = (343) / 0.42368
f = 171.5 / 0.42368
f = 404.79 Hz
Therefore, the lowest possible frequency of the emitted tone is 404.79 Hz
Answer:
2.57 seconds
Explanation:
The motion of the ball on the two axis is;
x(t) = Vo Cos θt
y(t) = h + Vo sin θt - 1/2gt²
Where; h is the initial height from which the ball was thrown.
Vo is the initial speed of the ball, 22 m/s , θ is the angle, 35° and g is the gravitational acceleration, 9.81 m/s²
We want to find the time t at which y(t) = h
Therefore;
y(t) = h + Vo sin θt - 1/2gt²
Whose solutions are, t = 0, at the beginning of the motion, and
t = 2 Vo sinθ/g
= (2 × 22 × sin 35°)/9.81
= 2.57 seconds
