A water balloon is launched with an initial speed of 40 m/s at 60 degrees above the horizontal. A tall building is 40 m from the launch site. Neglect air resistance and use g 10 m/s2 8. How much time does it take for the water balloon to hit the building? a) 2.0 s b) 2.4 s c) 2.8 s d) 3.2 s e) 3.6s ,D 9. At what height above ground level does the water balloon hit the building? a) 49.3 m b) 45.3 m c) 41.3 m d) 37.3 m e) 33.3 m 10. How fast is the water balloon moving as it hits the building? a) 33.8 m/s b) 30.8 m/s c) 27.8 m/s d) 24.8 m/s e) 21.8 m/s

a) 2.0 s is the time after which the balloon hit the building

a) 49.3 m is the height at which the balloon hits the building

d) 24.8 m/s is the velocity of the balloon at the instant of hitting the building

Explanation:

Given:

initial speed of projection of balloon,

angle of projection from the horizontal,

distance of a tall building from the launch site,

Now we find the horizontal component of velocity :

Now the time taken to hit the building:

(since we don't have any acceleration in the horizontal direction )

Now the vertical component of the velocity:

Time taken to reach the maximum height:

At max. height the final velocity (v=0)

(-ve, since the direction of velocity is opposite to the gravity)

Therefore the the balloon hit the building while ascending in height.

Now the height reached in the time in which the balloon hits the building:

(-ve, since the direction of velocity is opposite to the gravity)

Speed of the balloon while it hits the building:

(-ve, since the direction of velocity is opposite to the gravity)

Now the net resultant of the two components:

because it says use -2,0,and 2 for x, we can plug those numbers into the equation. y/4=3/4(-2)+2   y/4= -1 1/2 +2 now to get x by itself, we can multiply each side by 4. y=-6+8 which equals y=2 when x=-2. now you can do the same thing with 0 and 2 as x. you get y=8 when x=0 and y=14 when x=2.

x   |   y

-2 |   2

0   |   8

2   |   14

c. newton's third law of motion