The height reached in timelt by a particle. thrown upward and with a s

1.	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)
Answer» 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.

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)

Answer»

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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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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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)