Answer:
The first step is to calculate the initial velocity of the seal. We can use the impulse-momentum theorem, which states that the change in momentum of an object is equal to the impulse applied to it. The impulse is equal to the force multiplied by the time, so:
Impulse = Force x Time
Impulse = 600 N x 0.4 s = 240 Ns
The change in momentum of the seal is equal to its final momentum minus its initial momentum. We can assume that the seal was initially at rest, so its initial momentum is zero. The final momentum can be calculated using the formula for projectile motion:
Final Momentum = Mass x Velocity
Final Momentum = 30 kg x V
V = sqrt(2gh), where g is the acceleration due to gravity and h is the height the seal is launched from. We can assume that the height is zero, so h = 0.
V = sqrt(2gh) = sqrt(2 x 9.81 m/s^2 x 0 m) = 0 m/s
Final Momentum = Mass x Velocity = 30 kg x 0 m/s = 0 kg m/s
So the change in momentum is:
Change in Momentum = Final Momentum - Initial Momentum = 0 kg m/s - 0 kg m/s = 0 kg m/s
Now we can use the impulse-momentum theorem to find the final velocity of the seal:
Impulse = Force x Time = Mass x Change in Velocity
240 Ns = 30 kg x Change in Velocity
Change in Velocity = 8 m/s
The final velocity of the seal can be broken down into its horizontal and vertical components. The horizontal component is:
Vx = V cos(40) = 8 cos(40) = 6.11 m/s
The vertical component is:
Vy = V sin(40) = 8 sin(40) = 5.13 m/s
Now we can use the kinematic equations to find how far the seal travels horizontally before landing. We can assume that the seal lands at the same height it was launched from, so its final height is zero. The time it takes to reach this height can be found using the vertical component of the velocity and the acceleration due to gravity:
Vy = Voy + a*t
0 = 5.13 m/s - 9.81 m/s^2 * t
t = 0.52 s
The horizontal distance traveled can be found using the horizontal component of the velocity and the time of flight. The time of flight is twice the time it takes for the seal to reach its maximum height, which can be found using the vertical component of the velocity and the acceleration due to gravity:
Vy = Voy + a*t
0 = 5.13 m/s - 9.81 m/s^2 * t
t = 0.52 s
The time of flight is:
Time of Flight = 2 x t = 2 x 0.52 s = 1.04 s
Now we can find the horizontal distance traveled using the formula:
Horizontal Distance = Vx x Time of Flight
Horizontal Distance = 6.11 m/s x 1.04 s = 6.35 m
So the seal lands 6.35 meters from where it was launched.