Physics Homework?

Question

A fighter climbs with a speed of 120 m/s at an angle of 50degrees, to an altitude of 850m. At which point he drops a bomb directed at a bunker on the ground. If the bomb is to strike the bunker
a.What is the ground distance from the plane to the bunker?
b.How long do the ground troops have to evacuate the bunker?
c.What is the impact velocity of the bomb?

Answer 1

Well, I think I have the idea, look, when he drops the bomb, this bomb starts with a velocity, yes, the velocity of the fighter, why ?, because, the fighter has a speed, and is the inertia of a mass, right ?

So, let's find the velocity : The velocity of the bomb, has two components

Vx = 120*cos(50) = 77.13 m/s

Vy = 120*sin(50) = 91.9 m/s

The bomb at the first moment, goes up, and then goes down, let's consider a graphic, follo the steps :

Graph the hill, with the altitude, and at the top ot the hill, graph the movement of the bomb, it goes up, with : 91.9 m/s, and it's also moving in "x" direction, directly to the bunker.

Considering g = 9.8 m/s^2

We need this : the time that the bomb takes in the air, and the altitude the bomb reaches before it starts falling to the bunker.

First, let's find the altitude that the bomb reaches :

At the top final speed : Vf = 0

0 = 91.9^2 - 2*9.8*h

h = 430.8 meters

the total altitude that the bomb reaches : 430.8 + 850 :

H = 1280.8 meters

Let's find the time :

0 = 91.9 - 9.8*t

t = 9.3 seconds

the bomb reached that altitude, and now, is going to fall directly to the bunker.

During the 9.3 secondes, the bomb has been moving horizontally :

X = 9.3*Vx = 717.3 m

But the bomb is still on the air :

Total altitude = 1280.8 m

1280.8 = 9.8 t'^2 / 2

t' = 16.1 seconds

this is another time, is the time the bomb takes when it reached it's maximum altitude until it reaches the bunker.

Now using Vx, we can find :

16.1*Vx = X'

X' = 1241.8 meters

X' + X is answer to a)

X + X' = 1959.1 meters.

the time, well the time for answer b is : t +t'

total time = 16.1 + 9.3 = 25.4 seconds

for answer C, we must find the final velocity, we can use the final movement, since the bomb reached it's maximum altitude until it reaches the bunker :

Vf^2 = 2*9.8*(total altitude) = 2*9.8*1280.8 :

Vf = final speed = 158.4 m/s

Note : Good problem, I like solving it

* Another note : The other guys, I think you are missing the "y" component of the bombs velocity, It has two components, so the bomb, at the first moment goes up, and then it goes down.

Hope that might help you

Answer 2

I'll tell you how to get the answer:

Use right triangles to find the bomb's horizontal velocity (which will stay the same if there's no air resistance). Find out how much time it will take for the bomb to fall 850m (easy). In that time, find out the distance the bomb would travel horizontally, and that's the answer to "a". The time it took is the answer to "b". For the impact velocity, use Vf^2 = V0^2 + 2aX , where Vf is the final velocity, V0 is the initial velocity, a is acceleration, X is the distance traveled. You will need to do this for the vertical velocity and find its hypotenuse with the horizontal velocity to find the total velocity.

Good luck!

Answer 3

I think you are missing one element of the question.
The answer to a is given, but to b and c we need the velocity of launch of the bomb. If the bomb is slimpy let go from the plane, then the ground troops have:
g = 10m/s^2
h = 850m
t*10 = 850
t = 850/10 = 85, or 1min 25sec.

Answer 4

Do you not pay attention in class or something?