If a football player kicks a football so that it spends 3.0 seconds in the air and travels 50.0 m, then find the velocity and angle of the kick. (22m/s, 41 degrees) , I need step by step solutions. please
Let vx(t) and vy(t) be the horizontal and verticalcoordinates of velocity of the ball at time t.
There is a constant gravitation force F = mg acting on the ball during its fly. This force is directed to the ground and is opposite tovy.
Hence the ball moves uniformly along x-axis and withconstant acceleration -g along y-axis, so vx is constant: vx(t) = vx(0) and vy(t) = vy(0) -g*t. Thus the ball moved t=3 seconds and fly s=50.0 m, whenceits horizontal velocity, i.e. vx(t) is equal to vx = s/t = 50/3= 16.667 m/s
Further notice that the ball flies up the same time as itgoes down, and at the maximal point the time was 1.5 seconds and itsvelocity vy=0, whence vy(1.5) = vy(0)- 9.8 * 1.5 = 0 so, vy(0) = 9.8 *1.5 = 14.7 m/s. Hence the full initial velocity is v = sqrt(vx(0)^2+vy(0)^2) =sqrt(16.667^2+14.7^2) =sqrt(277.79+216.09) = sqrt(493.88) = 22.223 m/s
Let alpha be the angle of the kick. Then tan(alpha) =vy/vx = 14.7/16.667 = 0.88198 whence alpha =acrtan(0.88198)= 0.72277 rad = 41.412 degree