Use the model for projectile motion, assuming no air resistance.A baseball, hit above the ground, leaves the bat at an angle of 45 degrees and is caught by an outfielder 3 feet above the ground and 300 feet from home plate. What is the initial speed of the ball, and how high does it rise?

Answer:s=29.93m/sh=22.88mStep-by-step explanation:we must find the initial speed,  we will determine its position (x-y).x component [tex]s=0+v_{0}cos\alpha.t+0=v_{0}cos\alpha.t[/tex]y component [tex]h=0+v_{0}sin\alpha.t-\frac{1}{2}gt^{2}=v_{0}sin\alpha.t-\frac{1}{2}gt^{2}\\[/tex]  since the ball is caught at the same height then h=0[tex]h=v_{0}sin\alpha.t-\frac{1}{2}gt^{2}=0\\v_{0}sin\alpha.t-\frac{1}{2}gt^{2}=0;v_{0}sin\alpha.t=\frac{1}{2}gt^{2}\\t=\frac{2v_{0}sin\alpha}{g}\\[/tex]where t= flight time;[tex]s=v_{0}cos\alpha.t[/tex], replacing t:[tex]v_{0}=\sqrt{\frac{sg}{sin2\alpha}}[/tex][tex]s=v_{0}cos\alpha(\frac{2sin\alpha }{g})=\frac{v_{0} ^{2}2sin\alpha.cos\alpha}{g}=\frac{v_{0} ^{2}sin(2\alpha)}{g}]: the values ​​must be taken to the same units[tex]300ft*0.3048m/ft=91.44m[/tex][tex]v_{0}=\sqrt{\frac{91.44m*9.8\frac{m}{s^{2}}}{sin2(45)}}=\sqrt{895.112(\frac{m}{s} )^{2} }=29.93\frac{m}{s}[/tex]To calculate the height you should know that this is achieved when its component at y = 0[tex]v_{y}=v_{0}sin\alpha-gt=0;gt=v_{0}sin\alpha\\\\ t=\frac{v_{0}sin\alpha  }{g}\\h=v_{0}sin\alpha  .t-\frac{1}{2}gt^{2}[/tex]replacing t;[tex]h=v_{0}sin\alpha(\frac{v_{0}sin\alpha}{g})-\frac{1}{2}g(\frac{v_{0sin\alpha}}{g}) ^{2}\\[/tex]finally[tex]h=\frac{(v_{0}sin\alpha)^{2}}{2g}=\frac{(29.95*sin45)^{2}}{2*9.8}=22.88m[/tex]