A stone is thrown vertically upwards with a velocity of 30 m

A stone is thrown vertically upwards with a velocity of 30 m/s from the edge of a cliff 200 m above a river. The downward acceleration of is 9.8 m/s^2 Find the velocity equation of the stone. Find the position equation of the stone. Find the time to reach the highest point.

Solution

v^2 = u^2 + 2*a*s

a = -9.8 m/sec^2

u = initial velocity = 30 m/sec

v = final velocity = ?

v = sqrt (30^2 - 2*9.8*s) for upward direction motion

after reaching the highest point

v = 0 m/sec

v1 = new final velocity

g = +9.8 m/sec

v1 = sqrt (0^2 + 2*9.8*s)

v1 = sqrt (19.6*s) for downward motion

B.

position equation of the stone

s = u*t - 0.5*9.8*t^2

s = y0 - y

y0 = 200 m

s = 30*t - 0.5*9.8*t^2

y = 200 - 30*t + 0.5*9.8*t^2

after reaching max point

h = max point

s = 0*t + 0.5*9.8*t^2

s = h - y1

y1 = h - 0.5*9.8*t^2

C,

time to reach the highest point

v = u + a*t

a = -9.8

u = 30

v = 0 at highest point

t = (v - u)/a

t = 30/9.8 = 3.06 sec