A projectile is fired from ground level at time t = 0, at an angle theta with respect to the horizontal. It has an initial speed v_0. In this problem we are assuming that the ground is level. a) Find the time t_H it takes the projectile to reach its maximum height. Express t_H in terms of v_0, theta, and g (the magnitude of the acceleration due to gravity). b) Find t_R, the time at which the projectile hits the ground. I understood the first part ti be (v_0sin(theta))/g because it took the equation to V_y =v_0sin(theta) - gt and solved for t. Now when looking through my textbook, I found that the answer for part b is exactly the same as part a, except multiplied by 2. Can someone explain to me why its multiplied by 2? I figured since its the part after when v_y = 0 aka the peak of the ball, could you just add the t after with the t before the peak?