Question

Question 4 of 20: Select the best answer for the question 4. A baseball pitcher throws a baseball with an initial velocity of 128 foot per second at an angle of 20 to the horizontal. The ball leaves the pitcher's hand at a height of 4 feet. Find parametric equations that describe the motion of the ball as a function of time. How long is the ball in the air? When is the ball at its maximum height? What's the maximum height of the ball? O A. x 120 28;y-1037 1 B.x = 120.28: y = -16.37 25 33.948 OC.X 120.28t:y-16.3.784 2.642 3.954 set OD. X = 120.281: y = -16 13.784 23.586 feet

Answer 1

So=4 feet Vo = 128 feet per second usual fiecare vertical relocily Vy=128xsino = 128x Sinzoo 12840.3420 = 43.78 feet 1sec Vsco=128x COSO = 128x COS 20° = 128x0.9396 =120.280 feet 1 second y = vaxt =120.28xt Note accelaration done to gravity = -32 feet I see? y = £at² + vyt + 50 £x(-32)x+²+ 43.78€ tu =-16+²+43.782 +4 vertical relocity at time t dy वन -32t+43.78 At maximum height -327+43.78=0 t = 43.78 32 maximum height = 16+(1.3683²43.7881-368+4 = 33.g4 feet = 1.36880C

for calculating when ball will be in aur When y will again earth. y0 -16+²+43.787 +4=0 166²_43.787-420 t -(-43-781751-43.7632-4 -u xl 6xlu) 2x16 t = 2-82, t = -0.088 tat-0.088 so, t= 2.82800 summarizing Correct Ansuer-B a = 120.28t, y=-16t² +43.78+4 Time ball will be in air = 2.82 sec Ball will be in maximum height in 183688ec maximum height = 33.gu feet