Question

A stone is dropped from the top of a tower 200m high and at the same time another is projected velocity of 50ms. Find where and when the two will meet​

Answer 1

Answer:

Let the two stones meet at a distance of x mfrom the top of the tower, and 't' be the time taken. Let us assume the downward direction as positive.

For the stone that is dropped its initial velocity u=0ms

−1

; displacement s = x and acceleration = acceleration due to gravity (g).

Using s=ut+

2

1

at

2

,

wegetx=(0)t+

2

1

gt

2

→(1).

For the stone that is projected vertically upwards, its initial velocity, u=−50ms

−1

; displacement s =-(200-x) and acceleration a =g.

using s=ut+

2

1

at

2

,

weget−(200−x)=−50×t+1/2gt

2

200=50t−1/2gt

2

+x→(1)(2)

Fromtheequations(1)and(2)

wehave200=50t−1/2gt

2

+1/2gt

2

⇒200=50∴t=4s

Answer 2

$\huge\fcolorbox{black}{lime}{AnsweR:}$

Let the two stones meet at a distance of x m from the top of the tower, and 't' be the time taken. Let us assume the downward direction as positive.

For the stone that is dropped its initial velocity u= $\bold{0\:ms^-^1}$ displacement s = x and acceleration = acceleration due to gravity (g).

Using s= $\bold{ut+\frac{1}{2}at^2 }$

$\bold{we\:get\:x\:=0t+\frac{1}{2}gt^2 ...(1) }$

For the stone that is projected vertically upwards, its initial velocity, $\bold{u=-50\:ms^-^1}$

Displacement s =-(200-x) and acceleration a =g.

using s= $\bold{ut+\frac{1}{2}at^2 }$,

$\bold{we\:get\:-(200-x)=-50*t+\frac{1}{2}gt^2, }\\\\\bold{200=50t-\frac{1}{2}gt^2+x \:=\:(1)(2), }\\\\\bold{From\:the\:equations\:(1)\:and\:(2),}\\\\\bold{we\:have\:200=50t-\frac{1}{2} gt^2+\frac{1}{2}gt^2 }\\\\\bold{=200=50,}\\\\\bold{t=4s.}$