A ship maneuvers to within 2.47×103 m of an
island’s 1.76 × 103 m high mountain peak and
fires a projectile at an enemy ship 6.13 × 102
m on the other side of the peak, as illustrated.
The ship shoots the projectile with an initial
velocity of 2.54×102 m/s at an angle of 72.6
◦
.
The acceleration of gravity is 9.81 m/s
2
.
To find the time of flight of the projectile, we can use the kinematic equation:
Δy = v0sin(θ)t - (1/2)gt^2
Where:
Δy = the vertical distance covered by the projectile (the height of the mountain peak in this case)
v0 = initial velocity of the projectile (2.54×10^2 m/s)
θ = launch angle (72.6°)
t = time of flight
g = acceleration due to gravity (9.81 m/s^2)
First, let's calculate the vertical distance covered by the projectile (Δy):
Δy = 1.76 × 10^3 m
Now, we can plug in the values and solve for t:
1.76 × 10^3 = (2.54 × 10^2)sin(72.6°)t - (1/2)(9.81)t^2
Simplifying the equation:
1.76 × 10^3 = 2.54 × 10^2 * 0.9458 * t - 4.905t^2
Rearranging the equation:
4.905t^2 - (2.54 × 10^2 * 0.9458)t + 1.76 × 10^3 = 0
Now we can solve this quadratic equation for t using the quadratic formula:
t = (-b ± √(b^2 - 4ac)) / (2a)
Where:
a = 4.905
b = -2.54 × 10^2 * 0.9458
c = 1.76 × 10^3
Plugging in the values and solving for t:
t = [-(2.54 × 10^2 * 0.9458) ± √((2.54 × 10^2 * 0.9458)^2 - 4(4.905)(1.76 × 10^3))] / (2 * 4.905)
Calculating the values inside the square root:
t = [-(2.54 × 10^2 * 0.9458) ± √(1307.684 - 34759.52)] / 9.81
Simplifying further:
t = [-(2.54 × 10^2 * 0.9458) ± √(-33451.836)] / 9.81
Since the value inside the square root is negative, it means that the projectile never reaches the required height and falls short. Therefore, the time of flight is not possible to calculate using the given data.