Projectile Motion (1 Viewer)

[behindinfinity]

[FONT=&quot]A projectile has a range of 318.2 m and rises to a maximum of 60m.

How would I find the:
a) initial velocity of the projectile
b) initial horizontal velocity
c) initial vertical velocity

Thanks in advance to anyone who replies.[/FONT]

 

[kevin101]

At max height vertical velocity is 0.
Use the formula v squared = u squared + 2as.
0 = u squared + 2 (-9.8)60. where a =g=-9.8 and s=60
from there u find that vert. u=34.3m/s
the use v=u+at at max height where vert. velocity is 0
0 = 34.3-9.8t
t=3.5 sec
therefore, the time of flight is 2f=7sec
then use range=horizontal velocity times t
318.2 = horiz. velo. times 7
horiz. u = 45.5m/s
then draw your triangle and pythagorus to find that u=57m/s and the angle is 37 degrees above the horizontal. Use tan theta = 34.3/45.5 for angle.

Done