this can be solve using the formala of free fall
t = sqrt( 2y/ g)
where t is the time of fall
y is the height
g is the acceleration due to gravity
48.4 s = sqrt (2 (1.10e+02 m)/ g)
G = 0.0930 m/s2
The velocity at impact
V = sqrt(2gy)
= sqrt( 2 ( 0.0930 m/s2)( 1.10e+02 m)
V = 4.523 m/s
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The motion of the ball on the vertical axis is an accelerated motion, with acceleration

The following relationship holds for an uniformly accelerated motion:

where S is the distance covered, vf the final velocity and vi the initial velocity.
If we take the moment the ball reaches the maximum height (let's call this height h), then at this point of the motion the vertical velocity is zero:

So we can rewrite the equation as

from which we can isolate h

(1)
Now let's assume that

is the initial velocity of the first ball. The second ball has an initial velocity that is twice the one of the first ball:

. So the maximum height of the second ball is

(2)
Which is 4 times the height we found in (1). Therefore, the maximum height of ball 2 is 4 times the maximum height of ball 1.
Translating the first sentence into equation we get, t = k(1/h)
where t is time in seconds, k is the constant and h is the horsepower. Substituting
the values in the equation we have, 12s = k(1/200) we have a k = 2400 seconds –
hp. To get the time at 240 hp we use the equation above and the constant, we
get, t = (2400 seconds-hp)(1/240hp) t = 10seconds.