A hot-air balloon moves vertically downwards at a constant velocity of 1,2 m·s⁻¹ - NSC Physical Sciences - Question 3 - 2017 - Paper 1

No, the hot-air balloon is not in free fall because it is moving downwards at a constant velocity of 1.2 m·s⁻¹. Free fall occurs when an object is only under the influence of gravity, with no other forces acting upon it.

To find the time it takes for the ball to hit the ground, we can use the following kinematic equation:

$s = v_i t + \frac{1}{2} a t^2$

where:

• $s$ is the displacement, which is -22 m (downwards),
• $v_i$ is the initial velocity, which is 1.2 m·s⁻¹ (downwards),
• $a$ is the acceleration due to gravity, which is approximately 9.8 m·s⁻² (downwards).

Plugging in the values gives:

$-22 = (1.2)t + \frac{1}{2}(-9.8)t^2$

Rearranging gives:

$4.9t^2 - 1.2t - 22 = 0$

Using the quadratic formula:

$t = \frac{-b \pm \sqrt{b^2 - 4ac}}{2a}$

Substituting $a = 4.9$, $b = -1.2$, and $c = -22$ will yield the time taken to hit the ground.

When the ball bounces back, it reaches a height determined by its new initial velocity of 15 m·s⁻¹. To calculate the maximum height, we can use the formula:

$v^2 = u^2 + 2as$

where:

• $v$ is the final velocity (0 at the maximum height),
• $u$ is the initial velocity (15 m·s⁻¹),
• $a$ is the acceleration (which is -9.8 m·s⁻²), and
• $s$ is the height.

Setting up the equation:

$0 = (15)^2 + 2(-9.8)s$

Solving for $s$ will give us the height reached above the ground after the bounce, plus adding this height to the initial height of the ball when it hit the ground will give the total height above the ground.