Find the max height of a ball being thrown upwards

[ponas.j]

I am trying to solve the following problem (I am an engineer and my brother asked for help with his homework, so I am a bit embarrassed that I cannot solve this problem)

A ball is thrown upwards from the ground and after t seconds it reaches h height. t is influenced by h with the following function: [math]h = 24t - 5t^2[/math]. What is the heighest h_max it can reach?

My train of thought, since this question is only 1 point:
1) find the next time the height is at 0, because then it would have hit the ground. (t=4.8)
2) The movement is symmetric and parabolic, so in half the time the ball will have reached the maximum height. (t=2.4)
But it gives me the wrong answer

 

[Otis]

h = 24t − 5t^2
What is … h_max
But [t=2.4] gives the wrong answer

Hi ponas. Your result is correct for the t-coordinate of the parabola's vertex point. However, the exercise is asking for the h-coordinate of the vertex (i.e., the maximum height). What value did you calculate for h, when t=2.4?

h = 24t – 5t^2

… t is influenced by h …

Technically speaking, I would say t influences h (not the other way around) because t is the independent variable and h is the dependent variable. That is, the height of the ball depends upon how much time has elapsed since it was thrown.

[imath]\;[/imath]

 

[Steven G]

The problem did ask for an h value and you had a formula that converts t into h, yet your answer was a t value..
Having said that I think that you did a fine job thinking through the problem. 1st, you knew it was a parabola and then you knew many things about parabola. I think that you did fine.

Now if you brother is in calculus maybe he should use calculus. If you want the max value of a function, then you compute the derivative and solve for 0. Clearly, based on the image that you have for the function it will be a max point when the derivative is 0


h(t) = 24t-5t^2
h'(t) = 24 -10t
24 - 10t =0
12=5t
t = 12/5 or 2.4
h(2.4) = ???