# 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?

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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.

For this problem let's take the acceleration due to gravity as 10m/s^2 acting downwards. Now the initial speed of the stone is 40 m/s acting upwards.

For a body under acceleration, distance is give by u*t + (1/2)*a*t^2 where t is the time, a is the acceleration and u is the initial velocity.

After 3 s its height is 40*3 - (1/2)*10*9 = 75 m

After 4 s its height is 40*4 - (1/2)*10*16 = 80 m

after 5 s its height is 40*5 - (1/2)*10*25 = 75 m

After 8 s its height is 40*8 - (1/2)*10*64 = 0 m