An object of mass 3.17 kg is projected into the air at a 47.5° angle. It hits the ground 3.63 s later. What is its change in momentum while it is in the air? Ignore air resistance. Give answer in kg⋅⋅m/s. Do not enter unit. Enter a positive number for an upward direction or a negative number for a downward direction.

To solve this problem, we need to find the initial and final velocities of the object and then use the formula for change in momentum, which is final momentum minus initial momentum.

Step 1: Find the initial velocity The initial velocity can be found using the formula for the range of a projectile, which is (v^2 * sin(2*angle)) / g = time. We know the time (3.63 s) and the angle (47.5°), and g is the acceleration due to gravity (9.8 m/s^2). Solving for v gives us the initial velocity.

Step 2: Find the final velocity The final velocity can be found using the formula v = u + gt, where u is the initial vertical velocity, g is the acceleration due to gravity, and t is the time. The initial vertical velocity can be found by multiplying the initial velocity by sin(angle).

Step 3: Find the change in momentum The momentum of an object is given by the product of its mass and velocity. The change in momentum is then the final momentum minus the initial momentum.

Let's calculate:

Step 1: We rearrange the range formula to solve for v: v = sqrt((range * g) / sin(2angle)) Substituting the given values: v = sqrt((3.63 * 9.8) / sin(247.5)) = 14.96 m/s

Step 2: The initial vertical velocity u = v * sin(angle) = 14.96 * sin(47.5) = 11.16 m/s The final velocity v = u - gt = 11.16 - (9.8 * 3.63) = -24.62 m/s

Step 3: The initial momentum = mass * initial velocity = 3.17 * 14.96 = 47.42 kgm/s The final momentum = mass * final velocity = 3.17 * -24.62 = -78.04 kgm/s The change in momentum = final momentum - initial momentum = -78.04 - 47.42 = -125.46 kg*m/s

So, the change in momentum while the object is in the air is -125.46 kg*m/s. The negative sign indicates a downward direction.