Question

A boy standing on top of a building throws a small ball from a height of H₁ = 37.0 m. (See figure.) The ball leaves with a speed of 24.5 m/s, at an angle of 53.0 degrees from the horizontal, and lands on a building with a height of H₂ = 16.0 m. Calculate for how long the ball is in the air. (Neglect air friction, and use g = 9.81 m/s².)

Answer 1

the time in the air can be found by just considering vertical motion (y)
d_y = v_yi t + 1/2 g t^2
>>> d_y = 16.0 - 37.0 = -21.0 m (down)
>>> v_yi = v_i sin(angle) = 24.5 m/s * sin 53.0 = 19.56 m/s
>>> g = -9.81 m/s/s because I take up as positive
so
-21.0 = 19.56 t + 1/2 (-9.81) t^2

-21.0 = 19.56 t - 4.905 t^2
or
4.905 t^2 - 19.56 t - 21.0 = 0
solve this quadratic equation for t

I get 4.8 s (only the positive value makes sense in this case)