Question

A rock is launched at angle theta=53.2∘ above the horizontal from an altitude of ℎ=182 km with an initial speed ????0=1.61 km/s. What is the rock's speed when it reaches an altitude of ℎ/2? (Assume ???? is a constant 9.8 m/s2 and ignore air resistance.)

Answer 1

Answer:

The rock's final speed at the required altitude will be 42.24 m/s.

Explanation:

Let's start by finding the initial vertical speed.

Vertical Speed = 1.61 * Sin (53.2°)

Vertical Speed = 0.8 m/s

We want to know the speed of the rock when it is at an altitude of 91 km.

The total displacement of the rock from its starting position will thus be equal to -91 km

We can use this in the following equation:

$s=u*t+\frac{1}{2} (a*t^2)$

$-91=0.8*t+\frac{1}{2} (-9.8*t^2)$

t = 4.3918 seconds

Thus it takes 4.3918 seconds to reach the required altitude. We can now find the speed as follows:

$V=U+at$

$V=0.8+(-9.8)*(4.3918)$

$V = -42.24$

Thus the rock's final speed at the required altitude will be 42.24 m/s.

Answer 2

Answer:

V = 2092m/s = 2.09km/s, θ = 62.6°

Explanation:

Given

h = 182km = 182000

H = h/2 =91km = 91000m

Vo = 1.61km/s = 1610m/s

θ = 53.2°

g= 9.80m/s²

In order to find V at H = h/2 we need to find the time when H =h/2 which we can calculate from the equation below which relates the known terms H, h, Vo, θ, and g and the unknown t

H = h + (VoSinθ)t - 1/2×gt²

t = time taken to reach the height H

So

91000 = 182000 + (1610Sin53.2°)×t – 1/2×9.80× t²

0 = 182000 – 91000+ 1289.2t – 4.90t²

Rearranging the equation to the left side

4.90t² –1289.2t –91000 = 0

Solving by the use of the quadratic formula or the use of a calculator,

t = 320.96 and –57.86

Time cannot be negative so we discard the negative value.

t = 321s

V has both horizontal and vertical components which we can cal6as follows.

At this time the velocity can be calculated by

Vy = VoSinθ –gt

Vy = 1610sin53.2° – 9.8×321 = –1856.62m/s

So Vy = -1.86km/s

There is no acceleration along the horizontal so Vx is constant and doesn't change during the flight of the rock.

Vx = VoCosθ = 1610Cos53.2 = 964.43m/s

V = √(Vy² + Vx²) = √(-1856.62² + 964.43²)

V = 2092.2m/s

θ = tan‐¹(Vy/Vx) = tan‐¹(-1856.62/964.43)

θ = -62.6° (62.6° below the horizontal)