whe you say i didnt square the 125: i thought you were meant to leave the (125/0.7v)^2 as a squared number to be able to put into the formula?
-76564.5v + 128.7v - 1.8
(-128.7 ±√(128.7)^2 - 4 (-76564.5)(-1.8))/2(-76564.5)
(-128.7 ±√534700.71)/-153129
(-128.7 ±731.23)/-153129
that right so...
not sure what you are referring to with the x^2 tag (please explain :) )
3 - 1.2 = -4.9t^2 + vsin(46)t
divide both sides by t getting
1.8/t = -4.9t + vsin(46)
1.8/t + 4.9t = vsin(46)
and now i am stuck. yup, very stuck.
x-axis 125 = Vi x cos(46)t
t = 125/Vi x cos(46)
then do i sub that into
y=-\frac{1}{2}g t^2+v_{y,0}t ?
0 = -4.9t^2 + vsin(46)t
divide both sides by t getting
0 = -4.9t + vsin(46)
4.9t = vsin(46)
4.9 = vsin(46)/t
4.9 = 0.72v/125/0.7v
yeah i gather there is no acceleration in the x-axis and that there is the
-9.81 acceleration in the y-axis. but i can't seem to figure out firstly the velocity to reach the 125m and then secondly how to account for the 3m high wall.
i've had a read of some general initial velocity thread...
Homework Statement
A baseball leaving the bat at 46 degrees at a height of 1.2m from the ground clears a 3m high wall 125 meters from home plate. what is the initial velocity of the ball?
that's my problems i am have a reasonable amount of trouble with it. any help would be much appreciated.