The height reached in timelt by a particle. thrown upward and with a speed of time u is given by a h=ut-1/2gt^2. find the time taken to reach the mat.( g=.9.8m/s^2)
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Answer:
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.
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