A stone is thrown vertically up from the tower of height 25m with a speed of 20m/s time taken to reach the ground?

1 Answer
Oct 23, 2017

The stone will reach the ground in 5 seconds after being thrown upwards from the tower.

Explanation:

When the stone is thrown vertically upwards,

#u# = initial velocity =# 20#m/s

It will reach some height where its final velocity will become zero,

#v = 0#m/s

#a= -g = -10ms^(-2)# ----negative as it is going in direction against the gravitational pull. (Consider #g= 10 m/s^2# for ease of calculation).

Height = distance it travels upwards will be :

#s = (v^2 -u^2)/ (2a)#

# s = (0 - 20^2)/ (-2xx10)# m

#s = -400/-20 #m

#s = 20#m

Time taken to reach this distance will be :

# t = (v-u)/a #

# t = (0- 20)/-10 = 2#

# t = 2# seconds ----------let this be #t_1#

So it reaches #20#m from the peak of tower of height #25#m. So to reach to ground,the stone has to cover:

#s= 20+25= 45#m.

When it starts falling from #45#m above ground, its initial velocity is zero, so:

#u = 0#

We know it will come down with uniform acceleration

#a=g =10 ms^-2#

We know : #s = ut + 1/2 at^2#

# 45 = 0xxt + 1/2 xx10xxt^2#

# 45xx2/10 = t^2#

# cancel45^9xx2/cancel10^2 = t^2#

# 9xx cancel2/cancel2 = t^2#

#therefore t^2 = 9#, i.e. # t = 3# seconds-----Let this be #t_2#

Total time taken after throwing the stone upwards will be :

#t = t_1+ t_2#

#t = 2+3= 5#seconds

#therefore#The stone will reach the ground in 5 seconds after being thrown upwards from the tower.