Question

From the top of a building 75m, a bullet is fired at an angle of 45° above the horizontal with an initial velocity of 270 m/s. Find the time until it hits the ground.

Answer 1

 The initial vertical velocity is downward and the acceleration due to gravity will increase this downward velocity

$V_i = 270 cos 45^0 = 190.91 \frac{m}{sec}$

$d=v_it+\frac{1}{2}at^2$

$75 =190.91\times t+\frac{1}{2}\times 9.8\times t^2$

Changing to standard quadratic form yields

$4.9 t^2 + 190.91t-75 = 0$

This equation can be solved with the quadratic Equation:

 

t = 0.39 sec