Question

A particle is projected vertically upwards and its height 'h' and time 't' are

connected by, h=80t-16t^2. Find the greatest height attained and acceleration

at that time​

Answer 1

Answer:

put dh/ht=0

dh/dt=80-32t

dh/dt=0

t=80/32

d^2h/dt^2

-32=a

as second derivative less than 0 it is maxima point

so t=80/32 it attends maximum height

h=200-100=100m

Step-by-step explanation:

Answer 2

Step-by-step explanation:

for maximum and minimum dh/dt = 0

so

dh/dt = 80 - 32t

so

80 - 32t = 0

32t = 80

t = 2.5m

Now

hmax = 80*2.5 - 16*2.5^2

maximum hight = 200 - 100

maximum hight = 100unit

again diffrenciating with respect to x

a = d^2h/dt^2 = -32 unit