Question

A bullet is fired straight upward with the velocity of 98 m/s from the

top of a building 100 m high. Find: a.) its maximum height above the ground b.) how long is

it in the air? c.) the velocity when it reaches the ground.

Answer 1

(a) Let's first find the time that the bullet takes to reach its maximum height above the ground:

$$v=v_0+gt,$$ $$0=v_0+gt,$$ $$t=\dfrac{-v_0}{g}=\dfrac{-98\ \dfrac{m}{s}}{-9.8\ \dfrac{m}{s^2}}=10\ s.$$

Finally, we can find its maximum height above the ground from the kinematic equation:

$$y_{max}=y_0+v_0t+\dfrac{1}{2}gt^2,$$ $$y_{max}=100\ m+98\ \dfrac{m}{s}\cdot10\ s+\dfrac{1}{2}\cdot(-9.8\ \dfrac{m}{s^2})\cdot(10\ s)^2,$$ $$y_{max}=590\ m.$$

(b) We can find the time of the bullet in the air from the kinematic equation:

$$y=y_0+v_0t+\dfrac{1}{2}gt^2,$$ $$0=100+98t-4.9t^2,$$ $$4.9t^2-98t-100=0.$$

This quadratic equation has two roots: $t_1=20.97\ s$, $t_2=-0.97 s$. Since time can't be negative, the correct answer is $t=20.97\ s$.

(c) We can find the velocity when bullet reaches the ground from the kinematic equation:

$$v=v_0+gt,$$ $$v=98\ \dfrac{m}{s}+(-9.8\ \dfrac{m}{s^2})\cdot20.97\ s=-107.5\ \dfrac{m}{s}.$$

The sign minus means that the velocity of the bullet directed downward.