# 02 - Bullet fired from the top of a building

**Problem 02**

A bullet is fired at an initial velocity of 150 m/s and an angle of 56° at the top of a 120 m tall building. Neglecting air resistance, determine the following:

- The maximum height above the level ground that can be reached by the bullet.
- The time for the bullet to hit the ground.
- The velocity with which the bullet will hit the ground.

**Solution 02**

$v_{oy} = 150 \sin 56^\circ = 124.36 ~ \text{m/sec}$

$y_{max} = \dfrac{{v_{oy}}^2}{2g} = \dfrac{124.36^2}{2(9.81)}$

$y_{max} = 788.19 ~ \text{m}$

$H_{max} = 120 + y_{max} = 120 + 788.19$

$H_{max} = 908.19 ~ \text{m}$ *answer*

$y = v_{oy}t - \frac{1}{2}gt^2$

$-120 = 124.36t - \frac{1}{2}(9.81)t^2$

$4.905t^2 - 124.36t - 120 = 0$

$t = 26.284 ~ \text{sec}$ *answer*

$v_{Gx} = v_{ox} = 83.88 ~ \text{m/sec}$

$v_{Gy}^2 = v_{oy}^2 - 2gy$

$v_{Gy}^2 = 124.36^2 - 2(9.81)(-120)$

$v_{Gy} = 133.49 ~ \text{m/sec}$

$v_G = \sqrt{{v_{Gx}}^2 + {v_{Gy}}^2} = \sqrt{83.88^2 + 133.49^2}$

$v_G = 157.656 ~ \text{m/sec}$ *answer*

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