# A ball kicked from the ground (Projectile Motion)

1. Nov 14, 2009

### Chandasouk

1. The problem statement, all variables and given/known data

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?

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$$\Delta$$x

$$\Delta$$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 ?

2. Nov 14, 2009

### Delphi51

Your horizontal and vertical components are correct.
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. Nov 15, 2009

### Chandasouk

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 ?