# Homework Help: Components of velocity

1. Oct 12, 2007

### klm

A ball is thrown toward a cliff of height h with a speed of 30m/s and an angle of 60 degrees above horizontal. It lands on the edge of the cliff 4.0s later.

How high is the cliff?
do you just use the equation h=.5gt^2 ? and put it in -9.80=g and 4=t ?

2. Oct 12, 2007

### Staff: Mentor

You also need to consider the initial speed in the y-direction; here's the complete equation:
$$y = y_0 + v_{0y} t - (1/2) g t^2$$

Last edited: Oct 12, 2007
3. Oct 12, 2007

### klm

would this be correct: y0=26 , v0=30, g=-9.8 , t=4 ... so y=26+(30x4)- .5(-9.8)(4)^2 = 224.4 m ?

i got vy by doing v0sintheta= 26

4. Oct 12, 2007

### Staff: Mentor

No. In the formula I gave:
y0 is your initial position, which I presume is on the ground at height = 0
v0 (which I'll change to v0y) is the vertical component of the initial velocity, what you call v0sin(theta)
g = 9.8 m/s^2

To make it less confusing, I'll relabel v0 to be v0y in my equation.

5. Oct 12, 2007

### klm

ohh sorry, so y=0 + 26(4) - .5(9.8)(4)^2 = 25.6 m

6. Oct 12, 2007

### Staff: Mentor

Good!

7. Oct 12, 2007

### klm

thanks! can you help me with the next part too..! What was the maximum height of the ball?
i think the equation is just the same as the one you wrote, but just cut t=4 in half to get the peak height so t=2 ..so y= 26(2)-.5(9.8)(2)^2 = 32.4 m ?

8. Oct 12, 2007

### Staff: Mentor

No, you can't assume that the peak is at half the time. After all, it lands up on a cliff, so it spent more time rising than falling. (If it fell back down to the original height, then you'd be correct.)

Instead, use a velocity equation for the y-direction to figure out the time when it reaches maximum height. Hint: At the maximum height, what's the vertical speed?

9. Oct 12, 2007

### klm

um i think the vertical speed should be 0

10. Oct 12, 2007

### klm

so would it be okay to use the equation vfy= viy +ayT so 0= 26+9.8t so t= 2.65 and then cut then stick that time in that first equation you gave me?

11. Oct 12, 2007

### klm

so y= 26(2.65) -.5(9.8)(2.65)^2 =34 .4m ?

12. Oct 12, 2007

### klm

are the equations i used alright?

13. Oct 12, 2007

### Staff: Mentor

Very good!

14. Oct 12, 2007

### klm

thank you Doc Al! do you mind one more question, it will be the last one i promise! =)

15. Oct 12, 2007

### klm

What is the ball's impact speed?

i thought what you should do is find the final velocity in the x component and y component. and i thought that vfx= 15 since there is no acc in the x direction and vfy= -26 b/c of neg acc. and then i thought you should take the magnitude, but this does not work out to be right. do you know what i am doing wrong

wait i think i did this wrong, look at my next post please

Last edited: Oct 12, 2007
16. Oct 12, 2007

### klm

oh no actually should vfy= 65.2 because i tried the equation vfy=viy+ay x T so 26+(9.8x4)= 65.2
so do i do square root (15^2 + 65.2^2) = 66.9 ?

17. Oct 12, 2007

### Staff: Mentor

Careful here. ay = -9.8 m/s^2.

18. Oct 12, 2007

### klm

ohhh so should it be vfy= -13.2 and then do the sqaure root (13.2^2 +15^2 ) =19.98

19. Oct 12, 2007

### Staff: Mentor

Good!

20. Oct 12, 2007

### klm

thank you so much doc al!!