A ball is hit at an angle of 36 degrees to the horizontal from 14 feet above the ground. If it hits the ground 26 feet away (in horizontal distance), what was the initial velocity? Assume there is no air resistance.
i have posted the same question earlier and i got some steps to the answer .i am very thankful to you .But with that I am not able to reach the correct numerical answer which is 22.415 ft./sec
Vi*sin36= initial vertical velocity
hf=hi+Viv*t-1/2 g t^2
0=14+Vi*sin36*t-1/2 (32.1)t^2
now, in the horizontal, we know it traveled in the same time 26feet.
26=Vi*cos36*t or t=16/Vicos36
putting that into the vertical equation above,
0=14+Vi*sin36*26/ViCos36-1/2 32.1 *26^2/Vi^2 cos^2 36
0=14+tan36*16 -16.05*26^2/(Vi^2 cos^2 36)
ok, then do all the calculator work, solve for Vi.