apparently the ball hits the ground in 2.9 seconds. The horizontal speed of the ball is 66*.707 = 46.67 ft/s, so in those 2.9 seconds the ball travels
46.67*2.9 = 135.34 ft
The runner starts 63*3 = 189 feet away, so he must run 53.66 ft in 2.9 seconds, or 18.5 ft/s
football is kicked off with an initial speed of 66 ft/s at a projection angle of 45 degrees. A receiver on the goal line 63 yd away in the direction of the kick starts running to meet the ball at that instant. What must be his minimum speed (in feet/second) if he is to catch the ball before it hits the ground?
I found the x and y velocity's to be
66cos45=46.669 and 66sin45=46.669 and then solved for time using the y velocity by using Y=Yo+VoT+1/2(a)(t^2)
0=0+46.669-16.085t^2
t(46.669-16.085t)
t=0 and t=2.9sec
After this im not really sure what to solve for next.
1 answer