Godhelpme Posted October 16, 2019 Posted October 16, 2019 (edited) A body jumps with initial speed 10m/s under a 30 degree angle with the horizon. Air resistence shouldn't be taken into account and g=10m/s2. What will the tangential acceleration be after 0.5 seconds?? Edited October 16, 2019 by Godhelpme
swansont Posted October 16, 2019 Posted October 16, 2019 3 minutes ago, Godhelpme said: A body jumps with initial speed 10m/s. Air resistence shouldn't be taken into account and g=10m/s2. What will the tangential acceleration be after 0.5 seconds?? You need to get/provide more information. Tangential to what?
studiot Posted October 16, 2019 Posted October 16, 2019 (edited) I think you are talking about the projectile equations under gravity, but where is your attempt? Edited October 16, 2019 by studiot spelling
Country Boy Posted November 2, 2019 Posted November 2, 2019 The acceleration due to gravity is -g (which you are taking to be -10) vertically. Taking y to be positive upward and x positive to the right, the equations of motion are x= v_x t and y= v_y t- (g/2)t^2. The initial speed is 10 m/s at 30 degrees so v_x= 10(cos(30))= 10(1/2)= 5 m/s and v_y= 10(sin(30))= 10(sqrt(3)/2)= 5 sqrt(3) m/s which is about 8.7 m/s. So we have x= 5t and y= 8.7t- 5t^2. From x= 5t, t= x/5 so y= 8.7(x/5)- 5(x^2/25)= 1.74x- 0.2x^2. That is, of course, a parabola opening downward. The slope of the tangent line, at a given x, is 1.74- 0.4x. After 0.5 seconds, x= 5(0.5)= 2.5 and y= 8.7(0.5)- 5(0.25)= 5.6. The slope of the tangent line is 1.74- 0.4(2.5)= 0.74 so the tangent line to the trajectory at that time is y= 0.74(x- 2.5)+ 5.6. The tangential acceleration is the projection of the vector (0, -10) on that line.
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