Question

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Answer 1

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let initial speed be = u

time taken to reach the top most place be = t

accelaration = 9.8

so maximum height be

$h = ut - \frac{1}{2} \times 9.8 \times t^2$

$h = ut - 4.9t^2$

Answer 2

Let initial speed be = u

time taken to reach the top most place be = t

accelaration = 9.8

so maximum height be

h = ut - \frac{1}{2} \times 9.8 \times t^2h=ut−21×9.8×t2

h = ut - 4.9t^2h=ut−4.9t2

Hope for help