A ball kicked from the ground (Projectile Motion)

  • Thread starter Chandasouk
  • Start date
  • #1
165
0

Homework Statement



A ball is kicked from the ground at an angle of 65° and at a speed of 8 m/s. What is its speed when it reaches 1/2 of the maximum height along the path, assuming that air resistance is negligible?

My head hurts from this.

I made a diagram of this event by using a vector pointing diagonally with 65 degrees to the horizontal

I used trig to find the velocity at the x and y components

sin65 = y/8

Y = 7.25m/s

cos65=x/8

x = 3.38m/s

from here, i didn't know what to do. I used the equation V^2 = Vi^2+2a[tex]\Delta[/tex]x

[tex]\Delta[/tex]x = .58287751m

Then, to obtain the height

tang65=H/.58287751m

H = 1.25m

then half of that is .6249m

and uh....conservation of energy

Intial 1/2mVi^2 = Final mgh + 1/2mVf^2

You can cancel out the m's

1/2(8)^2 = (9.80)(.6249m) + .2Vf^2

Vf= 7.19 = 7.2m/s ?
 

Answers and Replies

  • #2
Delphi51
Homework Helper
3,407
11
Your horizontal and vertical components are correct.
Your max height is not.
Recommend you make two headings and write the appropriate equations:
Horizontal: constant speed so
-----------
d = vt

Vertical: accelerated so
--------
V = Vi + at, d = Vi*t + .5at^2
At maximum height, V = 0. Put that and your known values for the vertical Vi and acceleration into the first equation and solve for time of max height.
Put that time into the second equation to find the maximum height.
 
  • #3
165
0
Your horizontal and vertical components are correct.
Your max height is not.
Recommend you make two headings and write the appropriate equations:
Horizontal: constant speed so
-----------
d = vt

Vertical: accelerated so
--------
V = Vi + at, d = Vi*t + .5at^2
At maximum height, V = 0. Put that and your known values for the vertical Vi and acceleration into the first equation and solve for time of max height.
Put that time into the second equation to find the maximum height.


0 = 7.25ms+(-9.80m/s^2)t

-7.25m/s = -9.80m/s^2t

t=.74 seconds

d = (7.25)(.74) + .5(-9.80)(.74)^2

D=2.68m and half of that is 1.34m

then conservation of energy

Intial 1/2mVi^2 = Final mgh + 1/2mVf^2

You can cancel out the m's

1/2(8)^2 = (9.80)(1.34m) + 1/2Vf^2

Vf = 6.14m/s ?
 

Related Threads on A ball kicked from the ground (Projectile Motion)

  • Last Post
Replies
2
Views
1K
  • Last Post
Replies
3
Views
615
  • Last Post
Replies
2
Views
1K
  • Last Post
Replies
3
Views
2K
Replies
2
Views
3K
  • Last Post
Replies
10
Views
5K
  • Last Post
Replies
6
Views
1K
  • Last Post
Replies
3
Views
1K
Replies
19
Views
1K
Top