For the first question:
c) To find out how far in front of the truck the car is 10.0 seconds later, we need to calculate the distance covered by the car during the constant velocity phase.
The initial velocity of the car was 28 m/s and it maintained this velocity for 10.0 seconds, so we can use the equation d = vt, where d is the distance, v is the velocity, and t is the time.
Substituting in the values, we get d = (28 m/s)(10 s), which gives us 280 meters. Therefore, after 10.0 seconds, the car is 280 meters in front of the truck.
For the second question:
a) To find the ball's speed when it hits the ground, we need to calculate the final velocity.
We can use the equation v^2 = u^2 + 2as, where v is the final velocity, u is the initial velocity, a is the acceleration, and s is the displacement.
Since the ball is thrown upward, the acceleration due to gravity acts in the opposite direction, so we need to use a negative value for acceleration. Let's take upward as the positive direction, so the initial velocity (u) is 2.8 m/s, the acceleration (a) is -9.8 m/s^2, and the displacement (s) is -3.6 m (negative because the ball is moving upward).
Plugging these values into the equation, we get v^2 = (2.8 m/s)^2 + 2(-9.8 m/s^2)(-3.6 m).
Simplifying, we have v^2 = 7.84 m^2/s^2 + 70.56 m^2/s^2.
Adding the terms together, we get v^2 = 78.4 m^2/s^2.
Finally, taking the square root of both sides of the equation to find v, we have v = √(78.4 m^2/s^2). This gives us approximately 8.85 m/s. Therefore, the ball's speed when it hits the ground is approximately 8.85 m/s.
b) To determine how long after the first ball is thrown the second ball should be dropped from the same window so that both balls hit the ground at the same time, we need to find the time of flight for the first ball.
To calculate the time of flight, we can use the equation 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 final velocity is 0 m/s (when the ball hits the ground), the initial velocity (u) is 2.8 m/s, and the acceleration (a) is -9.8 m/s^2, we can rearrange the equation to solve for time.
0 = 2.8 m/s + (-9.8 m/s^2)t.
Simplifying, we have -9.8 m/s^2t = -2.8 m/s.
Dividing both sides of the equation by -9.8 m/s^2, we get t = (-2.8 m/s) / (-9.8 m/s^2).
Calculating this, we have t ≈ 0.286 seconds.
Therefore, if the second ball is dropped at the same time as the first ball is thrown, both balls will hit the ground at the same time.