The height reached in time t by a particle thrown upward with a speed u is given by h=ut-1/2gt^2. Find the time taken in reaching the maximum height.

Question

(Hint for h to be maximum, dh/dt=0)

Answer 1

 Without worrying about the derivative

dh/dt = 0,x this can be done by using

the quadratic.

The vertex is at (-b/2a, h(-b/2a))

 

h(t) = (-1/2)gt² + ut

 

a = -g/2, b = u

 

-b/2a = -u/(-g) = u/g

 

h(u/g) = (-1/2)g(u/g)²  + u²/g

= -u²/2g + u²/g

= (-u²+2u²)/(2g)

= u²/2g

 

The maximum height is u²/2g 

occurring at time t=u/g

 

Just another way to look at the problem,

arriving at the same answer as Patrick D.

 

 

 

Answer 2

 

Yes it is.

 

0 =  dH/dt = U - gT

 

 

  U = gT

 

  T = U/g = Tmax

 

So at that time , H(Tmax)

 = u(U/g) - 1/2(g)(U/g)^2

 

 

    = U^2/g - 1/2*U^2/g = U^2/(2g)

 

 

So it takes Tmax = U/g amount of time to reach the max height of U^2/(2g)

 

 

For grins and giggles, let's plug something in....

Let's say the speed is U=10 m/sec

 

Then it takes Tmax = 10/9.8 or just over 1 second to reach the 

 max height of 10^2/(2*9.8) or roughly 100/20 = 5 meters, which

is just a bit higher than 15 feet. Lots of rounding and estimate, but

it's just a sanity check.