A hot air balloon is rising upwards at a constant velocity of 5 m/s . When the balloon is 60 m above the ground , a sand bag is released from it and allowed to free fall . Ignore the effect of air resistance and assume that the balloon continues to move with the same constant velocity.

(a) Describe the motion of the sand bag when it was dropped to free fall.
(b)Where is the sand bag after 3 seconds?
(c)What is the maximum height reached by the sand bag .
(d)What is the magnitude and direction of the sandbag's velocity at the specific moment when it was released?
(e)How far apart will be the sandbag and the balloon be after 3 seconds?
(f)How long will it take the sandbag to reach the ground after it has been released?

maximum height reached by the sand bag is..change in y=(vf+vi/2).change in time..solution=(0+5/2)3=(2.5)(3)=7.5+60=67.5m

d).magnitude and direction of the sandbag's velocity..solution=vf=vi+a(change time)=0=5+(9.8)(3)=34.4 m/seconds

(a) When the sandbag is dropped from the hot air balloon, it will experience free fall motion. This means that it will accelerate downwards due to the force of gravity acting on it, while the balloon continues to move upwards with a constant velocity.

(b) After 3 seconds, the sandbag would have fallen a distance of h = (1/2)gt^2, where g is the acceleration due to gravity (approximately 9.8 m/s^2) and t is the time in seconds. Plugging in the values, we get h = (1/2)(9.8)(3)^2 = 44.1 m.

(c) The maximum height reached by the sandbag can be determined using the equation h = (1/2)gt^2. At the maximum height, the velocity of the sandbag will be zero. We can find the time it takes for the sandbag to reach this maximum height by setting the equation v = u + at to zero, where v is the final velocity (zero in this case), u is the initial velocity (5 m/s upwards), a is the acceleration due to gravity (-9.8 m/s^2 downwards), and t is the time. Solving for t, we get t = -u/a = -(5)/(-9.8) ā‰ˆ 0.51 seconds. Plugging this value into the equation for height, we get h = (1/2)(9.8)(0.51)^2 ā‰ˆ 1.26 m.

(d) The magnitude of the sandbag's velocity when it was released is 5 m/s downwards, as it is dropped from a balloon moving upwards at a constant velocity of 5 m/s.

(e) After 3 seconds, the balloon would have moved 3 * 5 = 15 meters upwards. The sandbag, on the other hand, would have fallen 44.1 meters downwards. Therefore, the distance between the sandbag and the balloon would be 44.1 + 15 = 59.1 meters.

(f) To determine the time it takes for the sandbag to reach the ground, we can use the equation h = (1/2)gt^2. Since the initial height is 60 meters, we set h = 60 and solve for t. Rearranging the equation, we get t = sqrt(2h/g). Plugging in the values, we get t = sqrt(2 * 60 / 9.8) ā‰ˆ 3.84 seconds. Therefore, it will take approximately 3.84 seconds for the sandbag to reach the ground after being released.

(a) When the sand bag is dropped from the hot air balloon, it will experience a free fall motion. Free fall means that the only force acting on the sand bag is gravity. Without considering air resistance, the sand bag will fall straight downwards in a vertical line.

(b) To determine where the sand bag is after 3 seconds, we can use the equation of motion for free fall. The equation is given as:

š‘¦ = š‘¦0 + š‘£0š‘” + (1/2)š‘Žš‘”^2

Here, š‘¦ is the final position (unknown), š‘¦0 is the initial position, š‘£0 is the initial velocity, š‘Ž is acceleration (due to gravity), and š‘” is the time taken.

Since the sand bag is dropped, š‘¦0 (initial position) would be 60 m above the ground, š‘£0 (initial velocity) is 0 m/s, š‘Ž (acceleration) is -9.8 m/s^2 (taking gravity to be 9.8 m/s^2 downward), and š‘” (time taken) is 3 seconds.

Substituting these values into the equation gives:

š‘¦ = 60 + 0 Ɨ 3 + (1/2) Ɨ (-9.8) Ɨ 3^2
= 60 + 0 + (-4.9) Ɨ 9
= 60 - 44.1
= 15.9 m

Therefore, the sand bag will be 15.9 m above the ground after 3 seconds.

(c) To find the maximum height reached by the sand bag, we need to calculate the time it takes for the sand bag to reach its highest point. At the highest point, the velocity of the sand bag will be zero.

Using the equation of motion:

š‘£ = š‘£0 + š‘Žš‘”

š‘£ is the final velocity, š‘£0 is the initial velocity, š‘Ž is acceleration (due to gravity), and š‘” is the time taken.

At the highest point, š‘£ = 0 m/s, š‘£0 = -9.8 m/s (initial velocity is downward), and š‘Ž = -9.8 m/s^2.

Substituting these values into the equation of motion gives:

0 = -9.8 + (-9.8) Ɨ š‘”

Solving for š‘”, we get:

9.8š‘” = -9.8
š‘” = -1

Since time cannot be negative, we discard this result. It means the sand bag will never reach a maximum height because it is falling straight down.

(d) When the sand bag is released, its initial velocity is the same as the velocity of the hot air balloon, which is upwards at a constant velocity of 5 m/s. However, as gravity acts on the sand bag, its velocity will change. At the moment it is released, the magnitude of its velocity is still 5 m/s, but its direction now changes to downward.

(e) Since the hot air balloon continues to move upwards at a constant velocity of 5 m/s and the sand bag falls freely, the distance between them will increase at the rate of their relative velocities. In this case, the relative velocity is the sum of the balloon's velocity and the sand bag's velocity. At release, the sand bag has a velocity of 5 m/s downward, while the balloon has a velocity of 5 m/s upward. So, the relative velocity is the sum of 5 m/s upward and 5 m/s downward, which is 0 m/s. Therefore, after 3 seconds, the sand bag and the balloon will still be 60 m apart.

(f) To find the time it takes for the sand bag to reach the ground after being released, we can use the equation of motion for free fall:

š‘¦ = š‘¦0 + š‘£0š‘” + (1/2)š‘Žš‘”^2

Here, š‘¦ is the final position (unknown - ground level), š‘¦0 is the initial position (60 m), š‘£0 is the initial velocity (0 m/s), š‘Ž is acceleration (due to gravity, -9.8 m/s^2), and š‘” is the time taken (unknown).

Since the sand bag reaches the ground, š‘¦ (final position) is 0 m. Substituting this value, š‘¦0, š‘£0, and š‘Ž into the equation of motion gives:

0 = 60 + 0 Ɨ š‘” + (1/2) Ɨ (-9.8) Ɨ š‘”^2

Simplifying the expression, we get:

0 = 60 - 4.9š‘”^2

Rearranging the equation, we have:

4.9š‘”^2 = 60

Dividing both sides by 4.9, we get:

š‘”^2 = 60/4.9

Taking the square root of both sides:

š‘” = āˆš(60/4.9)

Evaluating this expression, we find:

š‘” ā‰ˆ 3.84 seconds

Therefore, it will take approximately 3.84 seconds for the sand bag to reach the ground after being released.