Answer:
(a) t = 4.52 sec
(b) X = 1,156.49 m
Explanation:
Horizontal Launching
If an object is launched horizontally, its initial speed is zero in the y-coordinate and the horizontal component of the velocity [tex]v_o[/tex] remains the same in time. The distance x is computed as .
[tex]\displaystyle x=v_o.t[/tex]
(a)
The vertical component of the velocity [tex]v_y[/tex] starts from zero and gradually starts to increase due to the acceleration of gravity as follows
[tex]v_y=gt[/tex]
This means the vertical height is computed by
[tex]\displaystyle h=h_o-\frac{gt^2}{2}[/tex]
Where [tex]h_o[/tex] is the initial height. Our fighter bomber is 100 m high, so we can compute the time the bomb needs to reach the ground by solving the above equation for t knowing h=0
[tex]\displaystyle t=\sqrt{\frac{2h_o}{g}}[/tex]
[tex]\displaystyle t=\sqrt{\frac{2(100)}{9.8}}=4.52\ sec[/tex]
(b)
We now compute the horizontal distance knowing [tex]v_o=256\ m/s[/tex]
[tex]\displaystyle x=(256).(4.52)[/tex]
[tex]X=1,156.49\ m[/tex]
