A projectile motion problem with solution to be verified.

Question

In a projectile motion problem, the height of the object from the ground is given by $y=ut-1/2gt^2$, where $u$ is vertical component of initial velocity and $g$ is gravitational acceleration. The maximum value of $y$ is reached at a time given by?

MY ANSWER

$$\frac{dy}{dt}=u-gt$$

To find maxima $$u-gt=0$$ $$u=gt$$ $$t=\frac{u}{g}$$

Applying the values in original equation

$$y=\frac{u^2}{g}-\frac{u^2}{2g}$$ $$y=\frac{u^2}{2g}$$

The correct answer is $\frac{u}{g}$. What is the issue with my answer?

The problem is asking the time at which the maximum value of $y$ is reached. So you are correct. — Robert Z, Aug 17 '19 at 15:24

score 3 · Accepted Answer · answered Aug 17 '19 at 15:36

3

You've already realised the question is asking for the time, and that's been correctly calculated as $t= \frac ug$. However, you haven't actually shown that this time corresponds to a maximum height. For that you need the second derivative test. The second derivative $\frac {d^2y} {dt^2} = -g < 0$, which means that this value is indeed a maximum and not a minimum or an inflexion point. This is necessary for a complete solution.

answered Aug 17 '19 at 15:36

Deepak

26,801

Agreed, but it was an mcq, so I thought that could be neglected while solving. – Aditya Aug 17 '19 at 15:53

score 0 · Answer 2 · answered Aug 17 '19 at 15:24

0

I realize that the question asked was to find the time. That way the correct answer is $\frac ug$ I won’t be deleting this question for anyone who may need it in the future.

answered Aug 17 '19 at 15:24

Aditya

6,191

A projectile motion problem with solution to be verified.

2 Answers2