Question

A projectile is thrown at an angle ß with vertical. It reaches a maximum height
H. The time taken to reach the highest point of its path is ??
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Answer 1

Answer: √(2gH)/g

We have H = (v2cos2β)/2g)

Or, vcosβ = √(2gH)

And t = (vcosβ)/g

t = √(2gH)/g

Hoe this helps!