Question

From top of the building 160m high a ball is dropped and the same time a stone is
projected vertically upward from the ground with velocity of 40m/s. Find the time
when the ball and stone will meet.​

Answer 1

Explanation:

Let us consider the distance they meet at to be "x"

meters from the ground

initial velocity of ball. = 0 m/s

initial velocity of stone. = 40m/s

the second law of motion gives us

(160 - x)+x = (0+g*t^2/2) + (40t - g*t^2/2)

160 = 40t

t = 4s

x = 40t - g*t^2/2

= 160 - 9.8*16/2

= 81.6mg

they meet at 81.6 meters from the ground,

4 seconds after their initial launch

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Answer 2

From top of the building 160m high a ball is dropped and the same time a stone is projected vertically upward from the ground with velocity of 40m/s.

We have to find the time when the ball and stone meet.

initial velocity of the ball, u₁ = 0 [ ∵ the ball is dropped from the top of tower ]

initial velocity of stone , u₂ = 40 m/s

we should use relative formula,

$S_{rel}=S_0+u_{rel}t+\frac{1}{2}a_{rel}t^2$

here,

initial position of the ball relative to that of stone, S₀ = 160 - 0 = 160 m

final position of the ball relative to that of stone = 0 [ ∵ they meet at a specific time and that time, its relative position must be zero ]

velocity of the ball relative to that of stone = 0 - 40 = -40

acceleration of the ball relative to that of stone = g - g = 0

∴ 0 = 160 + -40t + 0

⇒t = 160/40 = 4sec

Therefore the time is 4 sec when the ball and stone meet