Quarterback passed a ball to a receiver while running at 1.5 m/s forward. The ba
ID: 2239334 • Letter: Q
Question
Quarterback passed a ball to a receiver while running at 1.5 m/s forward. The ball was released at a height of 2.5 meters with an angle of 45 degrees and an initial velocity of 20 m/s. The receiver ran from behind to catch the ball at a height of 2 meters. a) find the distance the ball traveled. b) Find the initial distance between the receiver and the quarterback, if the receiver ran at a constant velocity of 7 m/s to catch the ball. Assume he started running at the same time as the ball was thrown. c) Find the final velocity of the ball and its direction.Explanation / Answer
y=yo+voy*t + 1/2*a*t^(2) x=xo+vox*t Vb/e=Vb/q + Vq/e Vb/e= velocity of ball relative to the earth Vb/q= velocity of ball relative to quarterback Vb/e= velocity of quarterback relative to earth Vb/e-x= Vb/q-x + Vq/e-x Vb/e-x= 20cos(45) + 1.5= 15.64m/s Vb/e-y= Vb/q-y + Vq/e-y Vb/e-y = 20sn(45) + 0= 14.14m/s y=yo+voy*t + 1/2*a*t^(2) 2=2.5 + 14.14*t - 4.905*t^(2) (quadratic formula) t= 3.20 A) x= xo + vox*t x= 15.64*(3.2015)= 50.07m B) x= xo + vox*t 50.07=xo +7*(3.2015) Xo = 27.6675m C) Vy= Voy*t +a*t Vy= 14.12 - 9.81(3.2015) = -17.28m/s Vb/e= 23.29m/s ?= -47.8 South of East