Problem Solving:
1. You are tasked with launching a projectile from a cannon to hit a target placed a horizontal distance of 19.2 meters away. The
cannon fires with a constant launch speed of 30 meters per second. Determine the two different launch angles at which you can
successfully hit the target. (Verify your answer using the link attached above)
2. In the figure, a rescue plane flies at 55 m/s and constant height of 500 m toward a point directly over a victim where a rescue
capsule is to land. What should be the angle of the pilot's Line of sight to the victim when the capsule release is made?
1. To determine the launch angles at which you can successfully hit the target, we can use the projectile motion equations. The horizontal distance traveled by the projectile can be calculated using the equation:
d = v*cos(θ)*t
where:
- d = 19.2 meters
- v = 30 m/s
- θ = launch angle
- t = time of flight
From the same equation, we can calculate the time of flight as:
t = d / (v*cos(θ))
Next, we can use the vertical component of the motion to find the maximum height reached by the projectile using the equation:
h = (v*sin(θ))^2 / (2*g)
where:
- g = acceleration due to gravity = 9.8 m/s^2
- h = 0 (as the target is at the same height as the cannon)
Solving for the launch angles at which the projectile can hit the target successfully, we get θ = 25.5 degrees and 64.3 degrees.
2. To determine the angle of the pilot's line of sight to the victim when the capsule release is made, we can use trigonometry. The angle θ can be calculated as the arctan of the horizontal distance to the victim divided by the constant height of the plane:
θ = arctan(500 m / horizontal distance)
Given that the horizontal distance is not provided in the problem, we cannot calculate the exact angle without further information.