![A body is projected upwards with an initial velocity of 30 m/s. If the horizontal range of projection is 75 m, at what angle with the horizontal is the ball projected? [g = 10 ms-2]](/_next/image?url=https%3A%2F%2Fohijasnjvprzodcmxltt.supabase.co%2Fstorage%2Fv1%2Fobject%2Fpublic%2Fwhatsapp-images%2F1780323272602-c3d675e86e928255.png&w=3840&q=75)
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.
ScanSolve AI Answer
Step 1: Recall the formula for the horizontal range of a projectile. The horizontal range R of a projectile launched with initial velocity u at an angle to the horizontal is given by: R = (u^2 (2))/(g) Step 2: Substitute the given values into the formula. We are given R = 75 m, u = 30 ms^-1, and g = 10 ms^-2. 75 = ((30)^2 (2))/(10) Step 3: Simplify the equation. 75 = (900 (2))/(10) 75 = 90 (2) Step 4: Solve for (2). (2) = (75)/(90) (2) = (5)/(6) (2) ≈ 0.8333 Step 5: Find the value of 2. 2 = ((5)/(6)) 2 ≈ 56.44^ Step 6: Calculate the angle of projection, . = (56.44^)/(2) ≈ 28.22^ The angle with the horizontal at which the ball is projected is approximately 28.2^. 28.2^ Send me the next one 📸