a veclocity and acceleration problem

[deeaveragestan]

the questions: a ball is thrown upward with initial velocity v0 cm/s and reaches a maximum heigh of h m. how high would it have gone if its initial velocity was 2vo? how fast must it be thrown upward to achieve a maximum heigh of 2h m?


how would you approach this question in a calculus way?
i figured out the answers are (first part :4h) (second part:sqrt root of 2) simply just by pluggin in numbers, and using physics equations,
but i would really wanna know the calculus way of doin this. please and thank you

 

[Ishuda]

the questions: a ball is thrown upward with initial velocity v0 cm/s and reaches a maximum heigh of h m. how high would it have gone if its initial velocity was 2vo? how fast must it be thrown upward to achieve a maximum heigh of 2h m?


how would you approach this question in a calculus way?
i figured out the answers are (first part :4h) (second part:sqrt root of 2) simply just by pluggin in numbers, and using physics equations,
but i would really wanna know the calculus way of doin this. please and thank you

A relative extrema (min or max) for a function is when the derivative of that function is zero [assuming a continuous differentiable function of course].

Since the velocity is the derivative of the distance, the min/max of the distance occurs if the velocity is zero. Since the second derivative of the distance (the acceleration) under these circumstances is negative, the min/max is a max.

Thus, under the usual conditions, an initial velocity of v₀ gives
v(t) = v₀ - g t
d(t) = v₀ t - \(\displaystyle \frac{1}{2}\) g t²
since d is the integral of v and we assume a starting height of zero. So, find the time where the velocity is zero and use that time to find the maximum height h in terms of v₀ and g.

Now write the same equations in terms of an initial velocity v₁=2v₀. Since you know the formula for the maximum height in terms of the initial velocity and g, just substitute 2v₀ for v₁ to get h₁ in terms of h. Similarly with the other question.