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Question

1. 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 per second with an angle 30 degree with respect too original direction .the goal post is a distance of 40 m from him. find out whether the ball reach the goal post. 3. If an object is thrown horizontally with an initial speed 10 M per second from the top of the building of height hundred metre. what is horizontal distance covered by the particle? 4. An object is thrown with initial speed 5 M per second set an angle of projection 30 .what is the height and range reached by the particle?

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Solution

this is not good ask one question at a time now i will answer 2 question for your sake Final answer : Angle of Projection is 45° STEPS: 1) According to the question, Horizontal Range = 4 * (Maximum Height) ---(1) 2) We know that when prijectile is fired at angle ( a) ,then H(max) = u^2( sin^2(a)) / 2g Horizontal Range, R = u^2 sin(2a)/g where u is initial velocity. Using equation (1), u^2 sin(2a)/g = 4 * u^2sin^2a/2g ===> sin(2a)=2sin^2 (a) ===>2sina cos a =2sin^2 a ===>sina =cos a ====>a=15 hence a is not equal to 0 hence projection angle is 45 degree 3) This is the case of horizontal projectile. Given, initial velocity , u = 20 m/s height , h = 100m use the formula, Horizontal range , X = u√(2h/g) So, X = 20 × √{2 × 100/10} X = 20 × √20 m Hence, horizontal distance covered by object is 20√20 m

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