Question

A stone is allowed to fall from the top of a tower 100 m high and at the same time another stone is projected vertically upwards from the ground with a velocity of 25 m/s. Calculate when and where the two stones will meet.​

Answer 1

GIVEN:-

(stone,during free fall)

• $\rm{Initial\:Velocity=0}$

• $\rm{Acceleration=10m/s^-2}$

• $\rm{Distance\:travelled=x}$

• $\rm{Time\:taken=t}$

FORMULAE USED:-

• ${\boxed{\rm{S=ut+\dfrac{1}{2}\times{a}\times{t}^2}}}$

Therefore,

$\implies\rm{S=ut+\dfrac{1}{2}\times{a}\times{t}^2}$

$\implies\rm{x=0+\dfrac{1}{\cancel{2}}\times{\cancel{5}}\times{t}}$

$\implies\rm{x=5t^2}$..........eq.1

Now,

When stone is thrown Vertically upwards.

GIVEN:-

• $\rm{Initia\: velocity=25m/s^-1}$

• $\rm{Distance\: Travelled=100-x}$....eq2

• $\rm{Acceleration=10m/s^-2}$

• $\rm{Time\:Taken=t}$

FORMULAE USED:-

${\boxed{\rm{S=ut+\dfrac{1}{2}\times{a}\times{t}^2}}}$

Therefore,

$\implies\rm{S=ut+\dfrac{1}{2}\times{a}\times{t}^2}$

$\implies\rm{(100-x)=25t+\dfrac{1}{\cancel{2}}\times{10}\times{t}^2}$

$\implies\rm{x=100-25t+5t^2}$..........eq.3

From eq.1 and eq.3

$\implies\rm{5t^2=100-25t+5t^2}$

$\implies\rm{100=25t}$

$\implies\rm{t=\dfrac{\cancel{100}}{\cancel{25}}}$

$\implies\rm{t=4s}$

After 4 second two stones will meet.

From eq.1

$\implies\rm{x=5t^2}$

$\implies\rm{x=5(4)^2=80m}$

Now,

Put the value of x in eq.2

$\implies\rm{100-x}$

$\implies\rm{100-80}$

$\implies\rm{20m}$

The stones will meet each other at the height of 20m.

Answer 2

Answer:

let "t" = time after which both stones meet

"S" = distance travelled by the stone dropped from the top of tower

(100-S) = distance travelled by the projected stone.

◆ i) For stone dropped from the top of tower

-S = 0 + 1/2 (-10) t²

or, S = 5t²

◆ ii) For stone projected upward

(100 - S) = 25t + 1/2 (-10) t²

= 25t - 5t²

Adding i) and ii) , We get

100 = 25t

or t = 4 s

Therefore, Two stones will meet after 4 s.

◆ ¡¡¡) Put value of t = 4 s in Equation i) , we get

S = 5 × 16

= 80 m.

Thus , both the stone will meet at a distance of 80 m from the top of tower.