Respuesta :
Answer:
A rock is launched at an angle theta = θ = 53.2° above the horizontal from an altitude h = 182 km with an initial speed [tex]v_{0}[/tex] = 1.31 km/s. What is the rock's speed (m/s) when it reaches an altitude of h/2? (Assume g is a constant 9.8 m/[tex]s^{2}[/tex])
The speed of the rocks is 1870.74 m/s
Explanation:
The initial potential and kinetic energy of the system is equal to the final kinetic and potential according to the law of conservation of energy.
Initial (PE + KE) = Final (PE + KE)
P E = mgh
K.E = 1/2m
This implies that;
[tex]mgh_{0} + \frac{1}{2} mv_{0} ^{2} = mgh_{1} + \frac{1}{2} mv_{1} ^{2}[/tex] ................1
Where m is the mass;
[tex]h_{0}[/tex] is the initial height = 182 km = 182 km x 1000 = 182 000 m
g is acceleration due to gravity = 9.8 m/ [tex]s^{2}[/tex]
[tex]v_{0}[/tex] is the initial velocity = 1.31 km/s = 1.31 km/s x 1000 m/km = 1310 m/s
[tex]h_{1}[/tex] is the final height = h/2 = 182000/2 = 91000 m
[tex]v_{1}[/tex] is the final velocity;
Rearranging equation 1 to make [tex]v_{1}[/tex] the subject formula, we cancel out m;
[tex]gh_{0} + \frac{1}{2} v_{0} ^{2} = gh_{1} + \frac{1}{2} v_{1} ^{2}[/tex]
Substitution into the equation;
(9.8 x 182000) + (1310[tex])^{2}[/tex]/2 = (9.8 x 91000) + [tex]\frac{v_{1} ^{2} }{2}[/tex]
1783600 + 858050 = 891800 + [tex]\frac{v_{1} ^{2} }{2}[/tex]
2641650 - 891800 = [tex]\frac{v_{1} ^{2} }{2}[/tex]
[tex]\frac{v_{1} ^{2} }{2}[/tex] = 1749850
[tex]v_{1}^{2}[/tex] = 3499700
[tex]v_{1}[/tex] = [tex]\sqrt{3499700}[/tex]
[tex]v_{1}[/tex] = 1870.74 m/s
Therefore the rocks speed is 1870.74 m/s
Question is incomplete.
Complete question is;
A rock is launched at an angle theta = 53.2 degree above the horizontal from an altitude h = 182 km with an initial speed vo = 1.31 km/s. What is the rock's speed (m/s) when it reaches an altitude of h/2? (Assume g is a constant 9.8 m/s².)
Answer:
Rock's speed = V = 1870.72 m/s
Explanation:
From the question, we are given that;
h = 182km = 182,000m
H = h/2 = 91km = 91000m
Vo = 1.31km/s = 1310m/s
θ = 53.2°
g= 9.8 m/s²
We can find the final speed V at the altitude, H = h/2 by;
H = h + (VoSinθ)t - 1/2×gt²
Where,
t = time taken to reach the height H
Plugging in the relevant values;
91000 = 182000 + (1310Sin53.2°)×t – ((1/2) × 9.8 × t²)
0 = 182000 – 91000 + 1048.96t – 4.9t²
0 = 91000 + 1048.96t – 4.9t²
Thus,
4.90t² – 1048.96t –91000 = 0
By using quadratic equation, we'll obtain;
t = 280.32 seconds and –66.25 seconds
t cannot be negative and so we pick the positive one.
So, t = 280.32 s
Now, V has both horizontal and vertical components which we can calculate;
The vertical component Vy at t = 280.32s;
Vy = VoSinθ –gt
Vy = 1310sin53.2° – 9.8 × 280.32 = –1698.18 m/s
Now, for the horizontal component, Vx is constant and doesn't change during the flight of the rock.
Thus, acceleration is zero. So,
Vx = VoCosθ = 1310Cos53.2 = 784.72m/s
Now, the magnitude of the final velocity will be the resultant of Vy and Vx.
Thus,
V = √(Vy² + Vx²)
Plugging in the relevant values,
V =√(-1698.18² + 784.72²)
V =√3499600.7908
V = 1870.72 m/s
