Hello K.
asked • 03/30/23At what speed did the athlete need to shoot the pebble to hit the target
An athlete needs to hit a moving target that is launched vertically. The target launcher launches the target at 54km/h and is 27m away from the athlete. If the athlete is reacting after 280ms still shooting with the 35° angle, at what speed did the athlete need to shoot the pebble?
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1 Expert Answer
Note that the height of the target = (.28+Δt)*15.83 m where 15.83 m/s = velocity of target and Δt is the amount of time the pebble requires to reach the target.
Since the angle of inclination is 35 degree and the distance is 27m, the height of the collision must be 27tan(35)=18.99 m. Using the above find Δt=.920 seconds
Since the distance traveled (assume a straight line) forms the hypotenuese of a right triangle it is sqrt(272+18.992)=32.97m. Calculate velocity of pebble to be 32.97/.920=35.85 m/s
Taveen S.
Completely incorrect…
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04/09/23
Taveen S.
When analyzing this problem we are given that the pebble is shot 280ms = .28s after the target is launched. It is very important to understand that the pebbles time of flight will be exactly .28s less than that of the target, because it is shot after. If the target has already been in the air for .28s, it’s time of flight is greater by exactly .28s. NOW, we can set Δt1 = the time of flight for the pebble, Δt2 = the time of flight for the target. Therefor Δt2 = Δt2, Δt1 = Δt2 - (.28s). Now that we have our time we can actually solve for the time and the max height of the target, then do the same for the pebble and find the velocity which it would needed to be launched at in order to reach that certain height in that particular time frame. We can start by solving for the maximum height of the pebble by using the kinematic equation without time, given Vi = 54km/h x1000 = 54000m/h ÷60 ÷60 = 15 m/s, A = -9.8m/s^2 and and since at max height this would automatically imply Vf = 0m/s. If you plug those values into vF^2 =v0^2 +2aΔd. If you solve it correctly you should get that the max height is 11.47m. Then you would need to find the time of flight, you would need to plug the same exact values into the equation without displacement since you are trying to find time. This equation would be vF=v0 + at. If you solve this correctly you should get the time of flight of the target to be 1.53s. Now that we know Δt2 = 1.53s, we can find Δt1. We can plug Δt2 into Δt1 = Δt2 - (.28s). If you do this correctly it would just be 1.53 - .28, which is 1.25. This means that Δt1, time of flight of pebble = 1.25s and Δt2, time of flight of target = 1.53s. Since we know that the horizontal velocity, Vx of a projectile, ALWAYS REMAINS CONSTANT. And we know that the Range, horizontal distance = 27m, and now we know the vertical and horizontal time, because Δt1 is the same for both vertical and horizontal components. This means that the horizontal velocity, Vx = 27m ÷ 1.25s. Since there is no acceleration and the horizontal velocity must remain constant, we can say that Vx = horizontal distance ÷ Δt1. This would give us a horizontal velocity of the pebble, 21.58m/s. Now we would need to solve for the vertical component, vertical velocity. Since we want to know the velocity of when they intersect, the vertical displacement of both the pebble and target need to be the same, same for the time. Well technically the time is not the same but since it is shot .28s after, they cancel each other out, this remains correct. So now we know our Δt = 1.25s, Δd = 11.47m, a = -9.8m/s^2. We know this because the max height of the target is 11.47m after 1.53s, therefor the time will also be 1.53 with the addition of the delay because 1.25 + .28 = 1.53s. And we are trying to solve for the velocity that will make them hit so the vertical displacement must be the same = 11.47m. We are trying to solve for initial velocity so we would use the equation without final velocity. This equation would be, Δd = ViΔt + 1/2aΔt^2) You should get the answer to be about 15.3 m/s as the vertical initial velocity. Since we now have both components of a right angle triangle we can use the Pythagorean theorem to solve for the hypotenuse, which in this case is the velocity they are asking for. This equation is, c= sqrt(a^2+b^2.) When you plug both components in you should get sqrt(21.58^2+15.3^2), the final outcome and the final answer to this problem would be 26.48m/s. This means that in order for the 2 objects to intersect, the athlete would need to shoot the pebble at a speed of 26.48m/s. Final answer, Speed = 26.48m/s
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04/09/23
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Taveen S.
When analyzing this problem we are given that the pebble is shot 280ms = .28s after the target is launched. It is very important to understand that the pebbles time of flight will be exactly .28s less than that of the target, because it is shot after. If the target has already been in the air for .28s, it’s time of flight is greater by exactly .28s. NOW, we can set Δt1 = the time of flight for the pebble, Δt2 = the time of flight for the target. Therefor Δt2 = Δt2, Δt1 = Δt2 - (.28s). Now that we have our time we can actually solve for the time and the max height of the target, then do the same for the pebble and find the velocity which it would needed to be launched at in order to reach that certain height in that particular time frame. We can start by solving for the maximum height of the pebble by using the kinematic equation without time, given Vi = 54km/h x1000 = 54000m/h ÷60 ÷60 = 15 m/s, A = -9.8m/s^2 and and since at max height this would automatically imply Vf = 0m/s. If you plug those values into vF^2 =v0^2 +2aΔd. If you solve it correctly you should get that the max height is 11.47m. Then you would need to find the time of flight, you would need to plug the same exact values into the equation without displacement since you are trying to find time. This equation would be vF=v0 + at. If you solve this correctly you should get the time of flight of the target to be 1.53s. Now that we know Δt2 = 1.53s, we can find Δt1. We can plug Δt2 into Δt1 = Δt2 - (.28s). If you do this correctly it would just be 1.53 - .28, which is 1.25. This means that Δt1, time of flight of pebble = 1.25s and Δt2, time of flight of target = 1.53s. Since we know that the horizontal velocity, Vx of a projectile, ALWAYS REMAINS CONSTANT. And we know that the Range, horizontal distance = 27m, and now we know the vertical and horizontal time, because Δt1 is the same for both vertical and horizontal components. This means that the horizontal velocity, Vx = 27m ÷ 1.25s. Since there is no acceleration and the horizontal velocity must remain constant, we can say that Vx = horizontal distance ÷ Δt1. This would give us a horizontal velocity of the pebble, 21.58m/s. Now we would need to solve for the vertical component, vertical velocity. Since we want to know the velocity of when they intersect, the vertical displacement of both the pebble and target need to be the same, same for the time. Well technically the time is not the same but since it is shot .28s after, they cancel each other out, this remains correct. So now we know our Δt = 1.25s, Δd = 11.47m, a = -9.8m/s^2. We know this because the max height of the target is 11.47m after 1.53s, therefor the time will also be 1.53 with the addition of the delay because 1.25 + .28 = 1.53s. And we are trying to solve for the velocity that will make them hit so the vertical displacement must be the same = 11.47m. We are trying to solve for initial velocity so we would use the equation without final velocity. This equation would be, Δd = ViΔt + 1/2aΔt^2) You should get the answer to be about 15.3 m/s as the vertical initial velocity. Since we now have both components of a right angle triangle we can use the Pythagorean theorem to solve for the hypotenuse, which in this case is the velocity they are asking for. This equation is, c= sqrt(a^2+b^2.) When you plug both components in you should get sqrt(21.58^2+15.3^2), the final outcome and the final answer to this problem would be 26.48m/s. This means that in order for the 2 objects to intersect, the athlete would need to shoot the pebble at a speed of 26.48m/s. Final answer, Speed = 26.48m/s.04/09/23