Question

Consider one such aircraft flying with a velocity of 109 m/s at an angle θ below the horizontal. The aircraft releases a bomb when its altitude is 2.15 km above sea level. At the point when the bomb is released, the magnitude of the displacement from the bomb to its target at sea level is 2.48 km. What is the angle θ? (Give your answer in degrees. Consider θ to be a positive value, measured downward from the horizontal axis.)

Answer 1

Answer:

angle 31.2º

Explanation:

In the launch of projectiles, we can use the formula for the range, but let's be careful, this formula throws the body from the ground, rises and then falls to the same height.  In this case the body is at the point of maximum height and descending to sea level, so the magnitude of the distance we have corresponds to half the range

      R/2 = Vo2 sin 2θ/g  

      sin 2θ = R g / 2 Vo²

      sin 2θ = 2.15 * 1000 9.8 / 2 109²

      sin 2θ= 0.8867

      θ = sin-1 0.8867 /2

      θ=  31.2º