During one of the games, you were asked to punt for your football team. You kick

Question

During one of the games, you were asked to punt for your football team. You kicked the ball at an angle of 49° with a velocity of 26 m/s. If your punt goes straight down the field, determine the average velocity at which the running back of the opposing team standing at 72 m from you must run to catch the ball at the same height as you released it. Assume that the running back starts running as the ball leaves your foot and that the air resistance is negligible. (Enter the magnitude of the velocity.)

Explanation / Answer

SIMILAR PROBLEM During one of the games, you were asked to punt for your football team. You kicked the ball at an angle of 32° with a velocity of 24 m/s. If your punt goes straight down the field, determine the average velocity at which the running back of the opposing team standing at 72 m from you must run to catch the ball at the same height as you released it. Assume that the running back starts running as the ball leaves your foot and that the air resistance is negligible x(t)=X+Vt+.5at^2 The above equation can be used in the vertical direction to find time (which you will need to answer the overall question) setting the height the ball is released equal to the height it lands, you can assume both x(t) and X =0 V (initial velocity) in the y direction will be the y component of the velocity or 24sin(32) so V=12.71806m/s (up) acceleration is going to be 9.8m/s^2 down so you can solve for time -Vt=.5at^2 -V/.5a=t t=2.5955s Next, you will want to find the horizontal distance traveled by the ball. x(t) assume X=0 (initial position) initial velocity will be the x component of the velocity given V=24cos(32) =20.35315m/s acceleration (horizontally) is zero so x(t)=20.35315m/s(2.5955s) =52.8266m To answer the final question, The running back will have to run 72-52.8266 m (in your direction) in 2.5955 seconds since velocity is the change in position divided by the change in time, his average velocity must be 19.1734m/2.5955s which equals 7.38717 m/s

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