This physics problem tests your understanding of fundamental physical laws and their applications. The step-by-step solution below breaks down the problem using relevant equations and physical reasoning.
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Here is the solution to the projectile motion problem: Question 3: A particle is projected from a point O on horizontal ground with an initial speed U m/s at an angle to the horizontal. It passes through the point (40,10). Given that = 3/4, find: a) The value of U. b) The horizontal and the vertical components of the velocity when t = 2 s. State, with a reason, whether the particle is rising or falling at this instance. c) The total time for which the particle is at a height greater than 5 m. (take g=10 m/s^2) First, let's determine and from = 3/4. Using a right-angled triangle with opposite side 3 and adjacent side 4, the hypotenuse is sqrt(3^2+4^2) = sqrt(9+16) = sqrt(25) = 5. So, = (3)/(5) and = (4)/(5). The equations of motion for a projectile are: Horizontal displacement: x = (U ) t Vertical displacement: y = (U ) t - (1)/(2) g t^2 a) The value of U. The particle passes through the point (x,y) = (40,10). Substitute x=40, y=10, = 4/5, = 3/5, and g=10 m/s^2 into the equations: Step 1: Use the horizontal displacement equation. 40 = (U · (4)/(5)) t 40 = (4U)/(5) t 200 = 4Ut Ut = 50 (1) Step 2: Use the vertical displacement equation. 10 = (U · (3)/(5)) t - (1)/(2) (10) t^2 10 = (3U)/(5) t - 5t^2 (2) Step 3: Substitute Ut=50 from (1) into (2). 10 = (3)/(5) (50) - 5t^2 10 = 3 × 10 - 5t^2 10 = 30 - 5t^2 5t^2 = 30 - 10 5t^2 = 20 t^2 = 4 t = 2 s (since t > 0) Step 4: Substitute t=2 s back into equation (1) to find U. U(2) = 50 U = (50)/(2) U = 25 m/s The value of U is 25 m/s. b) The horizontal and the vertical components of the velocity when t = 2 s. State, with a reason, whether the particle is rising or falling at this instance. The velocity components are given by: Horizontal velocity: v_x = U Vertical velocity: v_y = U - gt Step 1: Calculate the horizontal velocity component at t=2 s. v_x = 25 · (4)/(5) v_x = 5 × 4 v_x = 20 m/s Step 2: Calculate the vertical velocity component at t=2 s. v_y = 25 · (3)/(5) - (10)(2) v_y = 5 × 3 - 20 v_y = 15 - 20 v_y = -5 m/s Step 3: Determine if the particle is rising or falling. Since the vertical component of velocity v_y = -5 m/s is negative, the particle is moving downwards. The horizontal component of velocity is 20 m/s. The vertical component of velocity is -5 m/s. The particle is falling because its vertical velocity component is negative. c) The total time for which the particle is at a height greater than 5 m. We need to find the times when y=5 m and then calculate the duration between these times. Using the vertical displacement equation: y = (U ) t - (1)/(2) g t^2 Substitute y=5, U=25, = 3/5, and g=10: 5 = (25 · (3)/(5)) t - (1)/(2) (10) t^2 5 = (5 × 3) t - 5t^2 5 = 15t - 5t^2 Step 1: Rearrange the equation into a quadratic form. 5t^2 - 15t + 5 = 0 Divide by 5: t^2 - 3t + 1 = 0 Step 2: Solve the quadratic equation for t using the quadratic formula t = -b ± sqrt(b^2 - 4ac)2a. Here, a=1, b=-3, c=1. t = -(-3) ± sqrt((-3)^2 - 4(1)(1))2(1) t = 3 ± sqrt(9 - 4)2 t = 3 ± sqrt(5)2 The two times when the particle is at a height of 5 m are: t_1 = 3 - sqrt(5)2 s t_2 = 3 + sqrt(5)2 s Step 3: Calculate the total time for which the particle is at a height greater than 5 m. This is the difference between t_2 and t_1. Total time = t_2 - t_1 = (3 + sqrt(5)2) - (3 - sqrt(5)2) Total time = 3 + sqrt(5) - 3 + sqrt(5)2 Total time = 2sqrt(5)2 Total time = sqrt(5) s The total time for which the particle is at a height greater than 5 m is sqrt(5) s.