Question

a stone is droped from a height of 400m high tower.at the same time another stone is projected vertically upward from the ground with a speed of 50m/s the stone will cross each other after a time . what is the time
​

Answer 1

Answer:

Total distance =400m

When first ball is dropped from  an hight :

u=0

Thrown upward:

u=50

t we have to find

x1=1/2gt^2

x2=50t-1/2gt^2

x1+x2=400

Evaluating the values

(1/2gt^2+50t-1/2gt^2)=400

50t=400

t=400/50

t=8s ans

Explanation:

Answer 2

Answer

The two stones will cross each other after 8 seconds

Given

a stone is dropped from a height of 400 m high tower.at the same time another stone is projected vertically upward from the ground with a speed of 50 m/s the stone will cross each other after a time

To Find

At what time they crosses each other

Concept Used

We need to apply equation of motion .

→ s = ut + ¹/₂ at²

Solution

Case - 1 : Object 1

Initial velocity , u = 0 m/s

[ ∵ dropped from height ]

Acceleration due to gravity , a = g m/s²

Distance , s = x₁ m

Time , t = t s

Apply 2nd equation of motion ,

$\to\ \rm s=ut+\dfrac{1}{2}at^2\\\\\to\ \rm x_1=(0)t+\dfrac{1}{2}gt^2\\\\\to\ \rm x_1=\dfrac{gt^2}{2}...(1)$

Case - 2 : Object 2

Initial velocity , u = 50 m/s

Acceleration due to gravity , a = - g m/s²

[ ∵ thrown against the gravity ]

Distance , s = x₂ m

Time , t = t s

Apply 2nd equation of motion ,

$\to\ \rm s=ut+\dfrac{1}{2}at^2\\\\\to\ \rm x_2=(50)t+\dfrac{1}{2}(-g)t^2\\\\\to\ \rm x_2=50t-\dfrac{gt^2}{2}...(2)$

Add (1) and (2) ,

$\to\ \rm x_1+x_2=\dfrac{gt^2}{2}+50t-\dfrac{gt^2}{2}\\\\\to\ \rm 400=50t\ [\; \because\ x_1+x_2=400\ m\ ]\\\\\to\ \rm 40=5t\\\\\to\ \rm t=8\ s\ \; \bigstar$

So , the two stones will cross each other after 8 seconds