A hot air balloon containing an AP Physics student ascends vertically at a constant speed of 8 m/s. While ascending, he accidentally drops his rubber duckie from the gondola of the ballong. Seven seconds after it is dropped the duckie bounces off a roof at point A and hits the ground at point B as shown in the sketch at the right.
A. Find the distance from the point where the duckie is dropped to point A
B. Find the speed of the sweet little rubber duckie just before it hits the roof
Suppose that immediately after bouncing off the roof, the velocity of the rubber duckie is 12m/s 37 degrees BELOW the horizontal and that point B is horizontal distance of 24 meters from point A.
C. Find how far point A is above point B
D. Find the duckie's time of flight between points A and B
E. Find the Velocity of the duckie the instant before it strikes the ground at Point B
I Put:
A. 240 meters
B. 68.6 meters
C. 49 meters
D. 2.51 seconds
E. 31.8 meters per second
To solve these problems, we need to use the equations of motion. Let's go through each part step by step.
A. To find the distance from the point where the duckie is dropped to point A, we can use the formula for the distance traveled in uniform motion:
distance = speed × time
Since the duckie was dropped and falls freely, the time it takes to reach point A will be the same as the time it took to fall. Using the formula:
distance = speed × time
distance = 8 m/s × 7 s
distance = 56 meters
Thus, the distance from the point where the duckie is dropped to point A is 56 meters.
B. To find the speed of the duckie just before it hits the roof (point A), we can use the equation of motion:
final velocity = initial velocity + acceleration × time
However, in this case, the acceleration due to gravity (9.8 m/s²) must be taken into account. Since the duckie is only influenced by gravity during its fall, the time it takes to reach the roof is the same as the time it took to fall (7 seconds).
Using the equation:
final velocity = initial velocity + acceleration × time
final velocity = 0 + 9.8 m/s² × 7 s
final velocity = 68.6 m/s
Thus, the speed of the duckie just before it hits the roof is 68.6 m/s.
C. To find how far point A is above point B, we need to consider the horizontal distance (24 meters) and the angle at which the duckie bounces off the roof (37 degrees below the horizontal). We can use trigonometry to find the vertical distance:
vertical distance = horizontal distance × tan(angle)
Using the formula:
vertical distance = 24 m × tan(37 degrees)
vertical distance = 24 m × 0.7536
vertical distance ≈ 18.086 meters
Thus, point A is approximately 18.086 meters above point B.
D. To find the duckie's time of flight between points A and B, we know that the horizontal distance is 24 meters. Since the motion is only in the horizontal direction, we can use the equation:
distance = speed × time
Substituting the known values:
24 meters = 12 m/s × time
time = 24 m / 12 m/s
time = 2 seconds
Thus, the duckie's time of flight between points A and B is 2 seconds.
E. To find the velocity of the duckie just before it strikes the ground at point B, we can again use the equation of motion:
final velocity = initial velocity + acceleration × time
Since the duckie only experiences the acceleration due to gravity during its fall, we can use the equation:
final velocity = 0 + 9.8 m/s² × 2 s
final velocity = 19.6 m/s
Thus, the velocity of the duckie just before it strikes the ground at point B is 19.6 m/s.