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A quarterback throws a football toward a receiver with an initial speed of $20 . \mathrm{m} / \mathrm{s}$ at an angle of $30 .$ above the horizontal. At that instant the receiver is $20 . \mathrm{m}$ from the quarterback. In (a) what direction and (b) with what constant speed should the receiver run in order to catch the football at the level at which it was thrown?

(a) In the direction of the motion of the ball.

(b) $v _ { x m c } = 7.52 \mathrm { ms } ^ { - 1 } .$

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could say that the time of flight is given by Delta y equaling V Y initial T plus 1/2 acceleration on why direction t squared. We can then say we know that this is equaling zero because Delta Wise equaling zero so we can say zero equals 20 meters per second. This would be multiplied by sine of 30 degrees multiplied by t and this would be plus 1/2 multiplied by negative 9.80 meters per second squared multiplied by t squared. And we find quickly this T will cancel out with one of these teas and we can see that tea will be equaling 2.0 seconds after algebraic Lee manipulating. So keep that answer in mind. Now the horizontal distance that the football moves in this time would be Delta X, our horizontal range. This would be equaling the ex initial times t, so this would be equaling again. There isn't any acceleration of acceleration vector in the ex direction. Therefore, we don't have to account for that acceleration term and so this would be equaling 20 meters per second multiplied by co sign of 30 degrees multiplied by 2.0 seconds and this is giving us 35 meters. So we can then say that four part eh, Because Delta X is equaling 35 meters and this is greater than 20 meters. This essentially means that the receiver must run away from the quarterback in the direction that the ball was thrown. And this is in order to catch the ball now for part B. Now the receiver has two seconds to run a distance of we can say d he used 20 meters away from the quarterback, so d is actually gonna be Delta X minus 20 meters. So this is equaling 35 minus 2015 meters. Therefore, his required speed would be equaling d over the T the time of 2.0 seconds because he has two seconds to run 15 meters essentially backwards or rather in the direction that the ball was thrown and so 15 meters divided by 2.0 seconds. Therefore, his speed must be 7.5 meters per second in the direction that the ball is thrown right when the quarter back throws the ball because then he has only two seconds. Um, he only has two seconds of air time essentially where he can actually run before he catches the ball. And so he must run to seven and 1/2 meters per second in order to catch the ball again 35 meters away from the quarterback. That is the end of the solution. Thank you for watching.