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2 years ago

What are the two most important characteristics of a geographical data set that must be known before any analysis can be done

Mathematics

1 answer:

Liula [17]2 years ago

3 0

The answer for this problem is b

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3 years ago

If a projectile is fired with an initial speed of vo ft/s at an angle α above the horizontal, then its position after t seconds

Elena L [17]

Answer:

a) The bullet hits the ground after 64 seconds.

b) The bullet hits the ground 113,511.7 feet away.

c) The maximum height attained by the bullet is of 16,384 feet.

Step-by-step explanation:

Equations of motion:

The equations of motion for the bullet are:

$x(t) = (v_0\cos{\alpha})t$

$y(t) = (v_0\sin{\alpha})t - 16t^2$

In which $v_0$ is the initial speed and $\alpha$ is the angle.

Initial speed of 2048 ft/s at an angle of 30o to the horizontal.

This means that $v_0 = 2048, \alpha = 30$ .

$x(t) = (v_0\cos{\alpha})t = (2048\cos{30})t = 1773.62t$

$y(t) = (v_0\sin{\alpha})t - 16t^2 = (2048\sin{30})t - 16t^2 = 1024t - 16t^2$

(a) After how many seconds will the bullet hit the ground?

It hits the ground when $y(t) = 0$ . So

$1024t - 16t^2 = 0$

$16t^2 - 1024t = 0$

$16t(t - 64) = 0$

16t = 0 -> t = 0 or t - 64 = 0 -> t = 64

The bullet hits the ground after 64 seconds.

(b) How far from the gun will the bullet hit the ground?

This is the horizontal distance, that is, the x value, x(64).

$x(64) = 1773.62(64) = 113511.7$

The bullet hits the ground 113,511.7 feet away.

(c) What is the maximum height attained by the bullet?

This is the value of y when it's derivative is 0.

We have that:

$y^{\prime}(t) = 1024 - 32t$

$1024 - 32t = 0$

$32t = 1024$

$t = \frac{1024}{32} = 32$

At this instant, the height is:

$y(32) = 1024(32) - 16(32)^2 = 16384$

The maximum height attained by the bullet is of 16,384 feet.

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3 years ago