A rock is launched at an angle θ = 53.2° above the horizontal from an altitude h = 182 km with an initial speed v0 = [04] km/s. What

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.)

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

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