# Question 04067

Jul 3, 2015

Here's how you can approach this problem.

#### Explanation:

FULL QUESTION

From a helicopter situated 600m of height from the earth, a bomb is slightly dropped to the ground. A soldier sees this in the very moment and he straight away fires a rifle bullet which has a velocity of ${\text{120 ms}}^{- 1}$.

• what is the time taken for the bullet to hit the bomb?
• when the two impacts what is the height to them from the ground?
• if the bullet rifle didn't hit the bomb, after another two seconds, he fires again in the same velocity. When it hits the bomb what is the height for the bomb from earth?

• if the bullet didn't hit the bomb as in the second question, after two seconds the soldier runs at a speed of ${\text{7 ms}}^{- 1}$. When the bomb fall on the ground how long can the soldier run?

!! LONG ANSWER !!

Let me start by saying that this is a very cool problem.

You know that the bomb starts falling from the helicopter. At the same time, a soldier that's located directly below the bomb, i.e. in the sport where the bomb will hit the ground, fires a bullet straight up.

The bullet and the bomb will presumably meet somewhere above gound level. The idea here is that if you add the distance travelled by the bomb and the distance travelled by the bullet in the same time $t$ you get 600 m, the height from which the bomd dropped.

${h}_{\text{bomb" + (-h_"bullet") = "600 m}}$

You need to account for the fact that the two objects travel in opposite directions, which means that their displacements will have opposite signs.

If you take the downward direction as the positive direction, then the distance travelled by the bomb will be

h_"bomb" = underbrace(v_0)_(color(blue)("=0")) * t + 1/2 * g * t^2

The distance travelled by the bullet will be

${h}_{\text{bullet}} = - {v}_{0} \cdot t + \frac{1}{2} \cdot g \cdot {t}^{2}$

This means that you have

$\frac{1}{2} \cdot g \cdot {t}^{2} - \left(- {v}_{0} \cdot t + \frac{1}{2} \cdot g \cdot {t}^{2}\right) = 600$

$\cancel{\frac{1}{2} \cdot g \cdot {t}^{2}} + {v}_{0} \cdot t - \cancel{\frac{1}{2} \cdot g \cdot {t}^{2}} = 600$

t = "600 m"/(120cancel("m")/"s") = color(green)("5 s")

The bomb and the bullet will heat each other after 5 seconds.

As you would image, the height from the ground will be equal for both objects, since they collide at the same point.

${h}_{\text{bomb" = 1/2 * 10"m"/cancel("s"^2) * 5^2 cancel("s"^2) = "125 m}}$

This is the distance the bomb travelled from its dropping point, which means that it will be

h_"above ground" = 600 - 125 = color(green)("475 m")

Now things get a little more complicated. If the bullet doesn't hit the bomb, then the soldier fires again, only this time he waits 2 seconds after seeing that he missed.

This means that, at the moment the second bullet is fired, the bomb travelled for

${t}_{\text{bomb" = 5 + 2 = "7 seconds}}$

Its height from the ground will be

${h}_{\text{bomb 2" = 600 - underbrace(1/2 * 10"m"/cancel("s"^2) * 7^2 cancel("s"^2))_(color(blue)(=h_2)) = "355 m}}$

At the moment the second bullet is fired, the bomb will have a velocity of

${v}^{2} = {\underbrace{{v}_{0}^{2}}}_{\textcolor{b l u e}{\text{=0}}} + 2 \cdot g \cdot {h}_{2}$

$v = \sqrt{2 \cdot 10 \cdot 245} = \text{70 m/s}$

The distance covered by the bomb after the second bullet is hit

h_"bomb" = underbrace(v_0)_(color(blue)("=70 m/s")) * t + 1/2 * g * t^2

The distance covered by the bullet

${h}_{\text{bullet}} = - {v}_{0} \cdot t + \frac{1}{2} \cdot g \cdot {t}^{2}$

This time, you have

h_"bomb" + (-h_"bullet") = 355

$70 \cdot t + \cancel{\frac{1}{2} \cdot g \cdot {t}^{2}} + 120 \cdot t - \cancel{\frac{1}{2} \cdot g \cdot {t}^{2}} = 355$

$t = \frac{355}{190} = \text{1.87 s}$

The bomb and the second bullet will meet 1.87 seconds after the second bullet is fired. The total falling time for the bomb will be

${t}_{\text{total" = 7 + 1.87 = "8.87 s}}$

The total distance it covered is

${h}_{\text{bomb total" = 1/2 * g * t_"total"^2 = 5"m"/cancel("s"^2) * 8.87""^2cancel("s"^2) = "393.4 m}}$

Its height from the ground at this point will be

h_"from ground" = 600 - 393.4 = color(green)("206.6 m")

Finally, two seconds after the second bullet misses the bomb, the soldier starts running. At this moment, the bomb will have travelled for

${t}_{\text{bomb total" = 8.87 + 2 = "10.87 s}}$

The time it takes the bomb to reach ground level is

$600 = \frac{1}{2} \cdot g \cdot {t}_{\text{ground}}^{2}$

${t}_{\text{ground" = sqrt((2 * 600)/10) = "10.95 s}}$

This means that the soldier has

t_"soldier" = 10.95 - 10.87 = color(green)("0.08 s")#

to get away.