You kick a football into the air. The vertical component of this motion can be modelled (...algebra question)?
The general form of the equation is S = s0 + v0t - .5at^2 Where s0=2 is the starting height, v0=31 ft/s is the initial velociy and a is the acceleration due to gravity, usually defined to be 32 ft/s^2 so you have S = 2 + 31t - .5(32)t^2 or S = 2 + 31t -16t^2 The football will have reached the ground when S = 0. 2 + 31t -16t^2 = 0 You can solve this with quadratic equation: t = (-31 +/- sqrt(31^2+4(2)16))/32 t = (-31 +/- sqrt(961+128))/(-32) = (-31 +/- sqrt(1089))/(-32) =(-31 +/- 33)/(-32) The positive answer is t = 2
A soccer ball is kicked with an initial horizontal velocity of 12 m/s and initial vertical velocity of 19 m/s?
1) Its horizontal velocity remains the same as 12 m/s. Its vertical velocity after 1.1 s = 19 - 9.81 * 1.1 m/s = 8.209 => its speed after 1.1 sec = √[(12)^2 + (8.209)^2] m/s = 14.54 m/s 2) Height above the ground after 1.1 s = 19 * 1.1 - (1/2) * 9.81 * (1.1)^2 m = 20.9 - 5.9 m = 15 m.
A kicked soccer ball has an initial velocity of 25 m/s at an angle of 40 ?
A kicked soccer ball has an initial velocity of 25 m/s at an angle of 40 above the horizontal level ground (neglect ir resistance) a- calculate the maximum height the ball reaches above its initial position. b- calculate the total time of the flight c- calculate the velocity of the projectile after 2 seconds. please pleaaaaaase someone help me with this am not good at physics am having a lot of problems with so please some one explain this to me BREIFLY!!
Math question: alexandria high school scored 37 points in a football game.?
This is a simultaneous equations problem, 3 equations and 3 variables. Let x be the number of touchdowns, y be the number of extra kicks and z be the number of two-point conversions. "The team scored 10 times during the game" tells us x + y + z = 10 "The team scored one fewer 2-point-conversions than extra kicks" tells us z = y - 1. Since 6 points are awarded for each touchdown, the number of points scored from touchdowns is 6x. One point is scored for each extra kick, so the number of points scored from extra kicks is y. Two points are scored for each two-point-conversion. So the number of points scored from the two-point conversions is 2z. 37 points are scored in total, so the third equation is 6x + y + 2z = 37. We have the three equations x + y + z = 10 z = y - 1 6x + y + 2z = 37. Substituting the 2nd equation into the other 2 gives x + y + y - 1 = 10 6x + y + 2(y - 1) = 37. Expanding out the brackets in the second equation gives x + y + y - 1 = 10 6x + y + 2y - 2 = 37. Simplifying both equations gives x + 2y = 11 6x + 3y = 39. Dividing the 2nd equation through by 3 gives x + 2y = 11 2x + y = 13. Rewrite the first equation as x = 11 - 2y. Substitute into the other equation to get 2(11 - 2y) + y = 13. Expanding out the bracket gives 22 - 4y + y = 13 Simplify -3y = -9 and therefore y = 3 Now x = 11 - 2y = 11 - 2*3 = 11 - 6 = 5 z = y - 1 = 3 - 1 = 2. So the team scores 5 otuchdowns, 3 extra kicks and 2 two-point conversions, and since the question asks how many touchdowns they scored the answer is 5.
The height of a soccer ball that is kicked...?
Do you know calculus, or just algebra? Let's do it without calculus. First find out when the ball is on the ground (height = 0) by finding the zeroes of the function. -16x² + 48x = 0 Factor out -16x: -16x(x - 3) = 0 Solutions: x = 0 or x = 3 That means the ball is on the ground when first kicked (time x = 0) and 3 seconds later (time x = 3). Because the shape is an upside down parabola which is symmetric, that means that the ball will be at its highest point halfway between 0 and 3. So take the time 1.5 and plug it back into the equation. y = -16(1.5)² + 48(1.5) y = 36 Answer: Maximum height = 36 feet Time to hit the ground = 3 seconds P.S. If you know calculus you can take the derivative and set it to zero. y' = -32x + 48 0 = -32x + 48 32x = 48 x = 48/32 x = 3/2 x = 1.5 seconds From there you can solve as before to get the maximum height of 36 feet.
I have to drop 30 seconds in the 100 meter butterfly and 5 seconds in the 50 meter freestyle in one year. How do I do this?
Swim faster.Alright, here are some tips, that are in no way a replacement for a coach!It won’t give you all the time off you want but do spend more time underwater. After your dive or turn, stay streamlined for 1–2 seconds longer than you think you should. Then start kicking and kick underwater for 1–2 seconds longer than you think you should. Then surface and swim without breathing for a stroke or two or five.In the 100 fly, look at the times you have done the fifty split in before. I suspect, without knowing you, that you are doing your first fifty in maybe 50 seconds and your second in 80 seconds. If you go five seconds slower in the first fifty, you will manage ten seconds faster in the second fifty. Again, I don’t know your numbers but they will work like the example above.Oh, and ask your coach for help!
In the NFL, can you try to kick a field goal on any down? What happens if you try on any down other than 4th and miss?
You can kick a field goal at any point you choose, but most people tend to wait until 4th down to attempt it, unless there are special circumstances (end of a half, for instance). There’s always a chance you could get the touchdown on an earlier down, or you can at least gain some yardage to make the field goal a bit easier.If the ball gets kicked past the line of scrimmage (whether it was tipped or not), then that is the end of the kicking team’s possession, regardless of down. Something similar to punt rules are employed when a missed FG lands in the field of play — the defending team can choose to return the miss as if it were a punt, or they can let it roll and get the ball at the spot where the kick was taken. The kicking team is not allowed to recover the miss unless the return team mishandles it.If it’s blocked backwards, however, then it’s a live ball. If the offense can manage to recover it, their possession would continue, as it does with any other play, with the next down. So, if they kicked on 3rd down and it was blocked backwards, they could recover the football and get another try at the kick on 4th down.
Physics Projectile Motion Problem?
Break the kick into two vectors: a horizontal and a vertical. The initial horizontal velocity is 12 cos 40 = 9.193 m/s The initial vertical velocity is 12 sin 40 = 7.713 m/s y = yo + vo*t + 1/2 a * t^2 0 = 0 + 7.713t - 4.9t^2 4.9t^2 = 7.713t 4.9t = 7.713 t = 1.574 seconds x = xo + vo*t + 1/2 a * t^2 x = 0 + 9.193(1.574) + 1/2 (0)(1.574)^2 x = 9.193(1.574) x = 14.471 meters 30 - 14.471 = 15.529 meters The ball lands 15.529 meters away from the second player after 1.574 seconds. distance = rate * time 15.529 = rate * 1.574 rate = 9.866 m/s Therefore, the second player must run at least 9.866 m/s.