How long is a projectile in the air for?

Question

Two physics students are playing catch with a baseball. One throws the ball to the other but overshoot and the ball lands on the library roof instead. The ball left the student's hand 2.08 m above the ground at a speed of 17.8 m/s and an angle of 64.5^o. The library is 8.08 m high and the ball lands on it while on its way down. How long was the ball in the air?

Answer 1

Solving for time directly requires solving a quadratic equation in time if the launch is asymmetrical for launch and landing (The solution for symmetrical launch is just 2v_y0/g = t_air)

Here you must solve y = y₀ + v_y0t - (1/2)gt²

8.08 m = 2.08m + (17.8 m/s * cos(64.5°))t - (1/2)(9.8 m/s²)t² (t that solves this is t_air)

The other way to solve this is to solve for v_y using

v_y² = v_yo² - 2g(y-y₀) then solve for t: t = (v_y - v_y0)/(-g)

This second way leads you through the quadratic equation without having to use it (at the cost of not solving for t directly)