Recent content by jarn6700

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...

1.8 = -4.9(125/0.7v)^2 + 0.72v(125/0.7v) 0 = -875v^2 + 128.57v -1.8is that correctly put into the equation? sorry i battle with quadratics...

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

this one too https://www.physicsforums.com/showthread.php?t=184043 but i don't get how they got to their conclusions.

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.