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 solutions to the problems: 4. A projectile is modeled by: h = -2t^2 + 12t + 5 a) Find the maximum height. Step 1: The maximum height occurs at the vertex of the parabola. The time t at the vertex is given by the formula t = -(b)/(2a). For h = -2t^2 + 12t + 5, we have a = -2 and b = 12. t = -(12)/(2(-2)) = -(12)/(-4) = 3 seconds Step 2: Substitute t=3 into the height equation to find the maximum height. h(3) = -2(3)^2 + 12(3) + 5 h(3) = -2(9) + 36 + 5 h(3) = -18 + 36 + 5 h(3) = 18 + 5 = 23 meters The maximum height is: 23 m b) Determine the time when the projectile hits the ground. Step 1: The projectile hits the ground when its height h is 0. Set the equation to 0. -2t^2 + 12t + 5 = 0 Step 2: Use the quadratic formula t = -b ± sqrt(b^2 - 4ac)2a. Here a = -2, b = 12, c = 5. t = -12 ± sqrt(12^2 - 4(-2)(5))2(-2) t = -12 ± sqrt(144 + 40)-4 t = -12 ± sqrt(184)-4 Step 3: Simplify sqrt(184). sqrt(184) = sqrt(4 × 46) = 2sqrt(46) t = -12 ± 2sqrt(46)-4 t = 2(-6 ± sqrt(46))-4 t = -6 ± sqrt(46)-2