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kiruha [24]
3 years ago
9

A missile is moving 1350 m/s at 25.0° angle. It needs to hit in a 55.0° direction in 10.20 s. What is the direction of its final

velocity?
Physics
1 answer:
77julia77 [94]3 years ago
8 0

Answer:

final velocity = 3504 m/s  

Explanation:

<em>Given data:</em>

velocity of missile = Vi = 1350m/s

angle at which missile is moving = 25degree

distance between missile and targets = 23500m

angle between target and missile=55degree

time=10.2s

<em>To find:</em>

Final velocity: ?

<em>Formula:</em>

x = Vx*t + ½*ax*t²  

Let x be the horizontal component of distance

x = ertical component of distance

t-time

ax = horizontal component of acceleration

ay = Vertical component of acceleration

Vx = horizontal component of velocity

Vy = Vertical component of velocity

<em>Solution:</em>

x = Vx*t + ½*ax*t²

23500m * cos55.0º = 1350m/s * cos25.0º * 10.20s + ½ * ax * (10.20s)²  

ax = 19.2 m/s²  

V'x = Vx + ax*t = 1350m/s * cos25.0º + 19.2m/s² * 10.20s = 1419 m/s  

<em>similarly vertically:</em>

y = Vy*t + ½*ay*t² 

23500m * sin55.0º = 1350m/s * sin25.0º * 10.20s + ½ * ay * (10.20s)²  

ay = 258 m/s²  

V'y = Vy + ay*t

     = 1350m/s * sin25.0º + 258m/s² * 10.20s = 3204 m/s  

V = √(V'x² + V'y²)

   = 3504 m/s  

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Answer:

Speed of the spacecraft right before the collision: \displaystyle \sqrt{\frac{2\, G\cdot M_\text{e}}{R\text{e}}}.

Assumption: the earth is exactly spherical with a uniform density.

Explanation:

This question could be solved using the conservation of energy.

The mechanical energy of this spacecraft is the sum of:

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  • the (gravitational) potential energy of this spacecraft.

Let m denote the mass of this spacecraft. At a distance of R from the center of the earth (with mass M_\text{e}), the gravitational potential energy (\mathrm{GPE}) of this spacecraft would be:

\displaystyle \text{GPE} = -\frac{G \cdot M_\text{e}\cdot m}{R}.

Initially, R (the denominator of this fraction) is infinitely large. Therefore, the initial value of \mathrm{GPE} will be infinitely close to zero.

On the other hand, the question states that the initial kinetic energy (\rm KE) of this spacecraft is also zero. Therefore, the initial mechanical energy of this spacecraft would be zero.

Right before the collision, the spacecraft would be very close to the surface of the earth. The distance R between the spacecraft and the center of the earth would be approximately equal to R_\text{e}, the radius of the earth.

The \mathrm{GPE} of the spacecraft at that moment would be:

\displaystyle \text{GPE} = -\frac{G \cdot M_\text{e}\cdot m}{R_\text{e}}.

Subtract this value from zero to find the loss in the \rm GPE of this spacecraft:

\begin{aligned}\text{GPE change} &= \text{Initial GPE} - \text{Final GPE} \\ &= 0 - \left(-\frac{G \cdot M_\text{e}\cdot m}{R_\text{e}}\right) = \frac{G \cdot M_\text{e}\cdot m}{R_\text{e}} \end{aligned}

Assume that gravitational pull is the only force on the spacecraft. The size of the loss in the \rm GPE of this spacecraft would be equal to the size of the gain in its \rm KE.

Therefore, right before collision, the \rm KE of this spacecraft would be:

\begin{aligned}& \text{Initial KE} + \text{KE change} \\ &= \text{Initial KE} + (-\text{GPE change}) \\ &= 0 + \frac{G \cdot M_\text{e}\cdot m}{R_\text{e}} \\ &= \frac{G \cdot M_\text{e}\cdot m}{R_\text{e}}\end{aligned}.

On the other hand, let v denote the speed of this spacecraft. The following equation that relates v\! and m to \rm KE:

\displaystyle \text{KE} = \frac{1}{2}\, m \cdot v^2.

Rearrange this equation to find an equation for v:

\displaystyle v = \sqrt{\frac{2\, \text{KE}}{m}}.

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\begin{aligned}v &= \sqrt{\frac{2\, \text{KE}}{m}} \\ &= \sqrt{\frac{2\, G\cdot M_\text{e} \cdot m}{R_\text{e}\cdot m}} = \sqrt{\frac{2\, G\cdot M_\text{e}}{R_\text{e}}}\end{aligned}.

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The mass of this bag of cement in S.I. units (kg) is equal to 0.062 kilograms.

<u>Given the following data:</u>

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To calculate the mass of this bag of cement in S.I. units (kg):

<h3>How to convert to S.I. units.</h3>

In Science, kilograms (kg) is the standard unit of measurement or S.I. units of the mass of a physical object. Thus, we would convert the value of the mass of this bag of cement in grams to kilograms (kg) as follows:

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