# Hardest 2 dimensional Kinematic Question I have ever encountered

Dooh
I just can't seem to figure this one out.

From the top of a tall building, a gun is fired. The bullet leaves the gun at a speed of 340m/s, parallel to the ground. The bullet puts a hole in a window of another building at "x" distance away, and hits a wall that faces the window that is "6.9m" away. The vertical distance in which the bullet traveled after it hits the window to the wall is "0.5m". The vertical distance between the gun on top of the roof and where the bullet strikes the wall is labeled as "y".

In this problem, i have to solve for both x and y. But with the given data, it seems impossible to solve for it.

• sathvik bellamkonda

Homework Helper
If you assumed that the bullet traveled on a straight line, you'd get a right angled triangle with x the base and y the height. Then x/y = 6.9/0.5. Now you need to produce a value for either x or y. Is this how you started thinking on this, too?

• Debkiran Mallick
stunner5000pt
very nice problem

work backwards

for hte time when the buller is in the building

Horizontal v1 = v2 = 340 m/s a=0 d = 6.9m t = ?
Vertical v1 = ? v2 = ? a = 9.8 m/s^2 d = 0.5m t=?
this time, t is the time the bullet is int he building ONLY
now what u need to do is find the time using whatever info u have above while NOT mixing X and y components. Time is the same, however

Now for the time when the bullet was between buildings
what is the intial VERTICAL velocity
what si the final vertical velocity (can you find it?)
for the time of flight for hte bullet between buildings
Horizontal v1 = v2 = 340 m/s a= 0 , d = x, t = ?
Vertical v1= you know v2 = you cna find out a = 9.8m/s^2 d = y-0.5 t = ?
you can solve for your unknowns now

• EnumaElish