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
Here are the calculations for the projectile motion problem. Given: Initial velocity u = 100 \, m/s Angle of projection = 30^ Acceleration due to gravity g = 10 \, m/s^2 a) The time at which the ball reaches the highest point (t_max) Step 1: Use the formula for the time to reach maximum height. The vertical component of velocity at the highest point is 0. t_max = (u )/(g) Step 2: Substitute the given values into the formula. t_max = 100 \, m/s × (30^)10 \, m/s^2 t_max = 100 \, m/s × 0.510 \, m/s^2 t_max = 50 \, m/s10 \, m/s^2 t_max = 5 \, s The time to reach the highest point is 5 s. b) The maximum height reached (H_max) Step 1: Use the formula for the maximum height. H_max = (u^2 ^2 )/(2g) Step 2: Substitute the given values into the formula. H_max = (100 \, m/s)^2 × ((30^))^22 × 10 \, m/s^2 H_max = 10000 \, m^2/s^2 × (0.5)^220 \, m/s^2 H_max = 10000 \, m^2/s^2 × 0.2520 \, m/s^2 H_max = 2500 \, m^2/s^220 \, m/s^2 H_max = 125 \, m The maximum height reached is 125 m. c) The horizontal range of the ball (R) Step 1: Use the formula for the horizontal range. R = (u^2 (2))/(g) Step 2: Substitute the given values into the formula. R = (100 \, m/s)^2 × (2 × 30^)10 \, m/s^2 R = 10000 \, m^2/s^2 × (60^)10 \, m/s^2 R = 10000 \, m^2/s^2 × sqrt(3)210 \, m/s^2 R = 10000 \, m^2/s^2 × 0.86602510 \, m/s^2 R = 1000 \, m × 0.866025 R ≈ 866.025 \, m The horizontal range of the ball is 866 m.