Refer to the diagram shown below.
u = 0, the initial vertical velocity
Assume g = 9.8 m/s² and ignore air resistance.
At the first stage of landing on the ground, the distance traveled is
h = 3.1 - 0.6 = 2.5 m.
If v = the vertical velocity at this stage, then
v² = u² + 2gh
v² = 2*(9.8 m/s²)*(2.5 m) = 49 (m/s)²
v = 7 m/s
At the second stage of landing on the ground, let a = the acceleration (actually deceleration) that his body provides to come to rest.
The distance traveled is 0.6 m.
Therefore
0 = (7 m/s)² + 2(a m/s²)*(0.6 m)
a = - 49/1.2 = - 40.833 m/s²
Answers:
(a) The velocity when the man first touches the ground is 7.0 m/s.
(b) The acceleration is -40.83 m/s² (deceleration of 40.83 m/s²) to come to rest within 0.6 m.
Final speed after collision with the wall

before collision the speed of ball initially

time taken for the collision

now as per the formula of acceleration we know that

now plug in all values in it


so acceleration is - 5 m/s/s for above situation
Answer:
Not understanding question
Explanation:
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Answer:
362.41 km/h
Explanation:
F = Force
m = Mass = 84 kg
g = Acceleration due to gravity = 9.81 m/s²
C = Drag coefficient = 0.8
ρ = Density of air = 1.21 kg/m³
A = Surface area = 0.04 m²
v = Terminal velocity
F = ma

Converting to km/h

The terminal velocity of the stone is 362.41 km/h