at a throwing speed pf 12.2 meters per seconds,what is the angle that produce a distance of 13 meters
under the influence of gravity?
consider the maximum height, velocity vertical is zero.
Vf=vi+gt
0=12sinTheta-9.8t
t=12sinTheta/9.8
time in the air is twice that time.
no, horizontal distance
13=12cosTheta*t=12cosTheta*2*12sinTheta/9.8
13=12^2/9.8 *2 cosTheta*sinTheta
remember siin2Theta=2sinTheta*cosTheta
solve for theta
13*9.8/144=sin 2Theta
check this.
Theta=50.6 deg