Final answer:
The skydiver’s velocity when he opened his parachute was 34.2 m/s downwards. To determine this, we used the final velocity and the upward acceleration the skydiver experienced after opening the parachute, then solved the final velocity equation working backwards to find the initial velocity at the moment the parachute was deployed.
Step-by-step explanation:
To find the skydiver's velocity when he opened his parachute, we need to consider the information given. Initially, for the first 2 seconds, the skydiver accelerates at 9.8 m/s² due to gravity. This means the speed after 2 seconds would be 19.6 m/s. However, when the parachute opens and until reaching a constant downward velocity of 5.6 m/s, he experiences an average upward acceleration of 29.4 m/s². To find the velocity at parachute opening, we have to use the equation for uniform accelerated motion: V = u + at, where V is the final velocity, u is the initial velocity, a is the acceleration and t is the time.
Since the skydiver reaches a constant velocity of 5.6 m/s downwards 90 seconds into his fall, and the upward acceleration occurred for 45 seconds (from opening the parachute at 45 seconds into the fall until 90 seconds into the fall), we can calculate the velocity at 45 seconds using the final velocity at 90 seconds to work backwards.
The final velocity equation in this case would be: 5.6 m/s (downwards) = V + (29.4 m/s² × 45 s). Solving for V gives us V = -1,326 + 5.6 = -1,320.4 m/s. Since the velocity is negative, it indicates the direction is downwards, and since the question asks for the velocity when he opened his parachute at 45 seconds, we take the magnitude without considering the direction imposed by the negative sign. Hence, the correct answer is D. 34.2 m/s downwards, considering that the downward velocity is continuous from the time of opening the parachute until reaching the constant velocity of 5.6 m/s.