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The height reached in a time t by a parcticle thrown upward with a speed is given by h=ut-1/2gt square where g=9.8m/secsquare is a constnt.Find the time taken in reaching the maxmimum height.

The height reached in a time t by a parcticle thrown upward with a speed is given by
h=ut-1/2gt square
where g=9.8m/secsquare is a constnt.Find the time taken in reaching the maxmimum height.

Grade:12

2 Answers

Saurabh Koranglekar
askIITians Faculty 10341 Points
one year ago
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

Vikas TU
14149 Points
one year ago
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
 
 

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