Question

An aeroplane is flying at a constant height of 1.96km with speed of 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

Answer 1

Dear Student,

◆ Answer -

θ = 30.41°

● Explanation -

# Given -

h = 1.96 km = 1960 m

v = 600 km/h = 167 m/s

# Solution -

Applying Newton's 2nd eqn of kinematics -

h = ut + 1/2 gt^2

1960 = 0×t + 1/2 × 9.8×t^2

t^2 = 1960 / 4.9

t^2 = 400

t = 20 s

So now, angle of inclination will be -

tanθ = h / vt

tanθ = 1960 / (167 × 20)

tanθ = 0.5868

θ = 30.41°

Therefore, angle of sight should be 30.41°.

Hope this helps you...