The horizontal range is four times
WebThe horizontal range is four times the maximum height attained by a projectile. The angle of projection is A 90∘ B 60∘ C 45∘ D 30∘ Solution The correct option is C 45∘ R =4 H cot θ, if R= 4H then cot θ= 1⇒ θ=45∘ Suggest Corrections 4 Similar questions Q. The horizontal range is four times the maximum height attained by a projectile. WebSep 4, 2024 · The horizontal range is four times the maximum height attained by a projectile. The angle of projection is#JEEMains #JEEAdvanced #NEET#IITJEEPhysicsLectures#...
The horizontal range is four times
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WebDec 20, 2024 · Prove that the maximum horizontal range is four times the maximum height attained by the project... - YouTube 0:00 / 2:22 Prove that the maximum horizontal range is four times the... WebJan 15, 2024 · (i) Prove that maximum horizontal range in four times the maximum height attained by projection. asked Jan 12, 2024 in Physics by Takshii ( 35.5k points) motion in a plane
WebJan 15, 2024 · Assertion : For angular projection, when angle of projection θ = tan-1(1), the horizontal range is four times the maximum height. Reason : The horizontal range of … WebVIDEO ANSWER: everyone in this question we are given that the horizontal range is four times of height maximum of a projectile so how can we like interfere this thing let there …
WebApr 5, 2024 · And R will be maximum, if sin 2 θ = 1 ⇒ 2 θ = 90 ∘ ⇒ θ = 45 ∘ Then, R max = u 2 g Therefore, the maximum height will be: H = u 2 sin 2 45 ∘ 2 g ⇒ H = u 2 g × 1 2 ⇒ H = R … WebApr 8, 2024 · The horizontal range is maximum when the angle of projection is 45°. So, Maximum height for angle of projection 45° is, Therefore, from equation (1) and (2), R = …
WebProve that the maximum horizontal range is 4 times the maximum height attained by a projectile. 890 Views Switch Flag Bookmark Advertisement Read each statement below carefully and state, with reasons and examples, if it is true or false: A scalar quantity is one that (a) is conserved in a process. (b) can never take negative values.
WebAug 24, 2024 · A particle is projected from the ground with some initial velocity making an angle of 0 45 with the horizontal. it reaches a height of 7.5m above the … ground while it travels a horizontal distance of 10m from the point of projection. find the initial speed of projection. ( 2 g m s 10 / ) lodge service station briton ferryWebAug 23, 2024 · The horizontal range is four times the maximum height attained by a projectile. 2,853 views Aug 22, 2024 48 Dislike Share Doubtnut 2.32M subscribers The horizontal range is four … individual health insurance nevada 2020WebOct 8, 2024 · A body is projected at such an angle that the horizontal range is three times the greatest height. The angle of projection is. asked Jan 15, 2024 in Physics by Sahilk (23.8k points) motion in a plane; aiims; neet +1 vote. 1 answer (A projectile projected at an angle `30^(@)` from the horizontal has a range R . If the angle of projection at the ... individual health insurance not aca compliantWeb1. An object at an angle such that the horizontal range is 4 times of the maximum height .what is the angle projection of the object? 2. Football player hits the ball with speed 20 m … individual health insurance oregon 2018WebThe answer is four times as much. Could someone explain? You are throwing a ball for the second time. If the ball leaves your hand with twice the velocity it had on the first throw, its horizontal range, compared to your first serve, would be what? The answer is four times as much. Could someone explain? Best Answer. This is the best answer ... lodges fallbarrowWebA: The equation for the horizontal range is given by Q: An arrow has an initial launch speed of 18 m/s. If it must strike a target 31m away at the same… A: initial sped = 18 m/s distance 31 m projection angle= ? Q: An arrow has an initial launch speed of 39 m/s If it must strike at a target 28 m away At the same… individual health insurance not obamacareWebFeb 1, 2024 · R_max = 4H_max The horizontal range is maximum for θ = 45° and it is given by R_max = (u^2)/g.....(1) And, the maximum height attained is H = (u^2(sinθ)^2)/(2g) And … individual health insurance oregon 2020