hey guys,

This physics homework is killing me and I need some help trying to do it

Equations of kinematics in two demensions and projetile motion

1) A volleyball is spiked so that it has an ititioal velocity of 15 meters per second (m/s) directed downwardat an angle of 55 degrees below the horizontal. What is the horizontal component of the balls velocity?

2)A rocket is fired at a speed of 75 meters per seond from ground level at an angle of 60 degrees above the horizontal. The rocket is fired toward an 11. 0 meter hight wall which is located 27 meters away. By how much does the rocket clear the top of the wall

Nonequilibrium applications of newtons second law

1. A 1380 Kg car is moving due east with an initial speed of 27 m/s. After 8.00 seconds the car has slowed down to 17.0 m/s. Find the magnitudeand direction of the net forcethat produces the deceleration

any help what so ever would be great

1) draw your triangle and use trig to solve.


2) i can't think right now, and i should know this but my brain isn't working. i apologize. firstly though, draw out your triangle and use trig to split up the components. i can tell you that much for now.

brain fart on the last one too. sorry.

well stephinika i thought about that too. I figured that since the angle opposite to the 55 is 35 and what i did was took cos(35)*15 should equal the length of the mystery angle. I got 12.3... Does that sound resaonable?

[edit]acutally the persons diagram is wrong let me do that again

^and youre wrong

sin35=vertical component/15
vertical component= 8.6m/s

ill edit with answers

2)A rocket is fired at a speed of 75 meters per seond from ground level at an angle of 60 degrees above the horizontal. The rocket is fired toward an 11. 0 meter hight wall which is located 27 meters away. By how much does the rocket clear the top of the wall

Ux= cos60 X 75
= 37.5m/s

Uy= sin60 X 75
= 64.9m/s

Ux is 37.5 therefore at the 25m mark the time is

1.5s

therefore

delta Y = Uyt + .5Ayt^2

delta Y = Uyt + .5(-9.8)t^2

delta Y = 64.9(1.5) - 4.9(1.5)^2

= 86.325

therefore since the wall if the wall is 11m tall the rocket clears it by 75.325m

edit with more

1. A 1380 Kg car is moving due east with an initial speed of 27 m/s. After 8.00 seconds the car has slowed down to 17.0 m/s. Find the magnitudeand direction of the net forcethat produces the deceleration
Fu=ma
= 1380 X 27
= 37260N

Fv = 1380 X 17
= 23460 N

difference 37260 - 23460
= 13800N to the west or -13800 to the east

im really not sure about this answer

hey guys i figured out the 1st and 3rd one and thanks for the help on the 2nd one wind

^ OK, I'm assuming you need no further help, if you do pm me to reopen this.

Topic Closed.