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s = vt

s = ut + ½ at^2

v = u + at

v^2 = u^2 + 2as

a = v/t and

a = (v-u)/t

And 1 sec = 1/3600 hr

So that, we can write,

25 m/s^2 = 25(0.001)(3600)^2 = 324000 km/h^2.

Initial velocity, u = 70 km/h = 70(1000)/3600 = 19.44 m/s

Final velocity, v = 0 m/s because ball will be stopped at the highest point.

Acceleration due to gravity, g = 9.8 m/s^2

We need to find the vertical distance that is height, h

And we can use the formula

v^2 = u^2 + 2as

0^2 = (19.44)^2 + 2(9.8)h

h = 19.28 m.

Velocity, v = 4.5 m/s

Time, t = 5 min = 5x60 = 300 sec

Now, we have the equation for displacement

s = vt

= 4.5(300) = 1350 m