A hot-air balloon moves vertically downwards at a constant velocity of 3.4 m·s⁻¹ - NSC Physical Sciences - Question 3 - 2024 - Paper 1

YES. The ball is in free fall because it is only subjected to the force of gravity, with no other forces acting on it during this time.

To find t₁, we can apply the equation of motion:

$v^2 = u^2 + 2a s$ Where:
v = final velocity,
u = 0 ext{ m·s}^-1 (at maximum height),
a = -9.8 ext{ m·s}^-2 (acceleration due to gravity),
s = -4.896 ext{ m} (the distance the ball rises).

Substituting values:

$0 = (7.2)^2 + 2(-9.8)(-4.896)$ This gives us t₁ = 1.44 s.

To find the height of the hot-air balloon, we use:
h = h_0 + v_0 t + rac{1}{2} a t^2 Where:
h₀ = 15 m,
v₀ = -3.4 m·s⁻¹,
a = -9.8 m·s⁻², and
t = 1.44 s.
Calculating, we get:

h = 15 - 3.4(1.44) + rac{1}{2}(-9.8)(1.44^2) \ = 15 - 4.896 \ = 10.1 m

For t₃, starting with the upward motion after the bounce:

Using the equation of motion:
v = u + at,
v = 0 ext{ m·s}^-1,
u = 7.2 ext{ m·s}^-1, and
a = -9.8 ext{ m·s}^-2.

Solving: $0 = 7.2 - 9.8 t_3$
t₃ = 0.73 s.
So, the total time from the release to maximum height is 1.44 + 0.73 = 2.17 s.