A stone is thrown upwards from a cliff top with a velocity of 40 m/s. What is its hight above the cliff-top after a) 3 s, b) 4 s, c) 5 s, d) 8 s?

The equation of motion of  the stone vertically thrown up
witha velocity of 40m/s  is given by y = ut-(1/2)gt^2, where u is the initial velocity
of and y is the vetical distance or height at any time t and g is the acceleration due
to gravity.

Give u = 40m/s . We presume g =
9.8m/s^2.

a) To find the height y of the stone from the
cliff-top after 3 seconds, put t = 3 sec , u = 40  g= 9.8 m/s^2 in y = ut-(1/2)gt^2
.

So y = 40*3-(1/2)9.8*3^2 = 75.9
meter.

b)Similarly to find the height when t = 4 sec, Put u
= 40, g= 9.8 and t = 4 in y = ut-(1/2) gt^2:

y = 40*4 -
(1/2)9.8*4^2 = 81.6 m

c) To determine height y when t = 5
seconds:

y = 40*5-(1/2)gt^2 = 77.5
m

d)

To determine height y
when t = 8 seconds.

y = 40*8- (1/2)g*8^2 =  6.4
m.