Answer
Pressure, P = 1 atm
air density, ρ = 1.3 kg/m³
a) height of the atmosphere when the density is constant
Pressure at sea level = 1 atm = 101300 Pa
we know
P = ρ g h


h = 7951.33 m
height of the atmosphere will be equal to 7951.33 m
b) when air density decreased linearly to zero.
at x = 0 air density = 0
at x= h ρ_l = ρ_sl
assuming density is zero at x - distance

now, Pressure at depth x


integrating both side


now,


h = 15902.67 m
height of the atmosphere is equal to 15902.67 m.
Explanation:
Given
initial velocity(v_0)=1.72 m/s

using 
Where v=final velocity (Here v=0)
u=initial velocity(1.72 m/s)
a=acceleration 
s=distance traveled

s=0.214 m
(b)time taken to travel 0.214 m
v=u+at


t=0.249 s
(c)Speed of the block at bottom

Here u=0 as it started coming downward

v=1.72 m/s
Answer:
0.1m/s²
Explanation:
Using the equation of motion
V = u+at
V is the final speed = 0.9m/s
Initial speed u = 0.5m/s
Time = 4secs
Get the acceleration
0.9 = 0 5+4a
0.9-0.5 = 4a
0.4 = 4a
a = 0.4/4
a = 0.1m/s²
Hence the acceleration is 0.1m/s²