Answer:
Part A
The height to which the projectile reaches is given by the following formula;

Where;
= The initial velocity of the projectile = Stays the same
y = The height the projectile reaches
y₀ = The height from which the projectile is launched
g = The acceleration due to gravity
t = The time taken
Given that the projectile has the same initial velocity, the variation of Δy with time, 't', will be the same when the project is launched from a different height, and the time change that it takes for the projectile to reach the maximum height will be the same for the previous launch height
Part B
From the change in eight of the projectile, Δy = y - y₀, the maximum height reached, 'y', is therefore given as follows;
y = Δy + y₀
The maximum height will increase by 'y₀', where 'y₀' is the difference between new height and the previous height, 
That is, if y₀ >
, the maximum height reached increases and if y₀ <
, y₀ will be added as y = Δy - y₀ and the maximum height reached decreases
Step-by-step explanation: