Answers:
a) -171.402 m/s
b) 17.49 s
c) 1700.99 m
Explanation:
We can solve this problem with the following equations:
[tex]y=y_{o}+V_{oy}t-\frac{1}{2}gt^{2}[/tex] (1)
[tex]x=V_{ox}t[/tex] (2)
[tex]V_{f}=V_{oy}-gt[/tex] (3)
Where:
[tex]y=0 m[/tex] is the bomb's final jeight
[tex]y_{o}=1.5 km \frac{1000 m}{1 km}=1500 m[/tex] is the bomb'e initial height
[tex]V_{oy}=0 m/s[/tex] is the bomb's initial vertical velocity, since the airplane was moving horizontally
[tex]t[/tex] is the time
[tex]g=9.8 m/s^{2}[/tex] is the acceleration due gravity
[tex]x[/tex] is the bomb's range
[tex]V_{ox}=350 \frac{km}{h} \frac{1000 m}{1 km} \frac{1 h}{3600 s}=97.22 m/s[/tex] is the bomb's initial horizontal velocity
[tex]V_{f}[/tex] is the bomb's fina velocity
Knowing this, let's begin with the answers:
With the conditions given above, equation (1) is now written as:
[tex]y_{o}=\frac{1}{2}gt^{2}[/tex] (4)
Isolating [tex]t[/tex]:
[tex]t=\sqrt{\frac{2 y_{o}}{g}}[/tex] (5)
[tex]t=\sqrt{\frac{2 (1500 m)}{9.8 m/s^{2}}}[/tex] (6)
[tex]t=17.49 s[/tex] (7)
Since [tex]V_{oy}=0 m/s[/tex], equation (3) is written as:
[tex]V_{f}=-gt[/tex] (8)
[tex]V_{f}=-(97.22)(17.49 s)[/tex] (9)
[tex]V_{f}=-171.402 m/s[/tex] (10) The negative sign ony indicates the direction is downwards
Substituting (7) in (2):
[tex]x=(97.22 m/s)(17.49 s)[/tex] (11)
[tex]x=1700.99 m[/tex] (12)
