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

The rock's speed when it reaches h/2 is ; 42.24 m/s

Given data :

∅ = 53.2°

Height ( altitude ) = 182 km

Initial horizontal speed = 1.61 km/s

g = 9.8 m/s²

First step ; Determine the vertical speed

Vertical speed ( u ) = 1.61 * sin ( 53.2°) = 0.8 m/s

next step : Determine the time to complete the total displacement of the rock from the starting point

s = 182 / 2 = 91 km,  a = 9.8 m/s²

s = ut + 1/2 ( a*t² ) ------ ( 1 )

- 91  = 0.8 * t + 1/2 ( 9.81 * t² )

∴ t = 4.40 secs

Final step : Determine the Final Rock speed when it reaches 91 km

V = U + at

   = 0.8 + ( 9.81 * 4.40 )

   = 42.24 m/s .

Hence we can conclude that The rock's speed when it reaches h/2 is ; 42.24 m/s

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