An aeroplane is flying at constant height of 1.96 km with speed 600 km/h towards a point directly over a target. At what angle of sight should
it release a bomb if it is to strike the target on the ground?
(1) 75°
(2) 45°
(3) 0°
(4) 60°
Answers
Hi,
Answer: option (4) 60°
Explanation:
Let’s assume that the aeroplane is flying at a height “h” from the ground at point A to point C i.e., from the figure attached below we can see h = AC = 1.96 km = 1960 m.
Also, let the point of target on the ground be at a point B from point A.
The speed of the aeroplane, Vx = 600 km/hr = 600 *(5/18) = 166.67 m/s
Step 1:
We will first calculate the time required by using the following equation,
s = ut + ½ gt²
⇒ h = 0 + ½ *9.8*t² …… [here s=h and initial speed of plane is zero]
⇒ 1960 = 4.9 t²
⇒ t = √400 = 20 s
Step 2:
Now we have to find the value of distance AB(refer to the fig. below).
∴ Distance, AB = speed(Vx) * time(t) = 166.67 * 20 = 3333.4 m
Step 3:
Finally we will find the angle of sight from where the aeroplane should release a bomb if it is to strike the target on the ground i.e., referring to the figure we will have to calculate the value for tan(∠C).
∴ tan (∠C) = perpendicular/ base = 3333.4/1960 = 1.70
∴ ∠C = tan⁻¹(1.70) = 59.53° ≈ 60°
Thus, in order to strike the target with a bomb on the ground, the angle of sight should be 60°.
Hope this is helpful!!!!
Answer:
See the attachment
Explanation:
