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?
The fact that t_flight equals to 2*t_peak (time taken for particle to reach its peak height) is usually made in simple projectile motion problems (When particle is dropped after flight to the same level as it was thrown) I guess you can prove it yourself by finding the expressions for its peak height. Then only thing you should do is to consider the particle's starting point is that height and it is thrown to the ground horizontally with a velocity of v0*cos(theta).