Question

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 is given by the parametric equations x=(v0cos(α))t andy=(v0sin(α))t−16t2
(where x and y are measured in feet).
Suppose a gun fires a bullet into the air with an Initial speed of 2048 ft/s at an angle of 30 o to the horizontal.
(a) After how many seconds will the bullet hit the ground?
(b) How far from the gun will the bullet hit the ground? (Round your answer to one decimal place.)
(c) What is the maximum height attained by the bullet? (Round your answer to one decimal place.)

Answer 1

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 $v_0 = 2048, \alpha = 30$.

So

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

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

(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.