# from the top of a tower 100m high a ball is dropped and at the same time

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## From the top of a tower of 100m height, a ball is dropped and at the same time another ball is projected vertically

From the top of a tower of 100m height, a ball is dropped and at the same time another ball is projected ... two balls will meet. Take g = 9.8 ms-2.

## From the top of a tower of 100m height, a ball is dropped and at the same time another ball is projected vertically

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## From the top of a tower 100 m in height a ball is dropped and at the same time another ball is projected vertically upwards from the ground with velocity of 25 ms^

Click here👆to get an answer to your question ✍️ From the top of a tower 100 m in height a ball is dropped and at the same time another ball is projected vertically upwards from the ground with velocity of 25 ms^-1 . Find when and where the two balls will meet. Take g = 10 ms^-2

From the top of a tower 100m in height a ball is dropped and at the same time another ball is projected vertically upwards from the ground with velocity of 25msQuestion −1

. Find when and where the two balls will meet. Take g=10ms

−2 Medium Open in App

Updated on : 2022-09-05

Solution Verified by Toppr

Step 1: Analysing the problem [Refer Figure]

Let the two ball meet at point C at time t

If the distance covered by ball 2 is x, then the distance covered by ball 1 will be 100−x, Since the total hieght is 100m.

Step 2: Equations of motion for constant acceleration

Since acceleration is constant, Therefore we can apply equations of motion for both balls (considering Positive Downwards).

For ball 1: u 1 =0 a 1 =g=10m/s 2 s 1 =100−x S 1 =u 1 t+ 2 1 a 1 t 2 ⇒ 100−x= 2 1 (10)t 2 ....(1) For ball 2: u 1 =−25m/s a 2 =g=10m/s 2 s 2 =−x S 2 =u 2 t+ 2 1 a 2 t 2 ⇒ −x=−25t+ 2 1 10t 2 .

⇒ x=−25t−5t

2 ....(2)

Step 3: Solving equations

From equation (1) and (2)

100−25t+5t 2 =5t 2 ⇒ t=4s

Putting the value of t in equation (1)

x=100−5×(4) 2 =20m

Hence they will meet 20m above the ground after t=4s

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## Form the top of a tower 100 m in height a ball is dropped and at the same time another ball is projected vertically upwards from the ground with a velcoity of 25 ms^(

Let (A) be the top of a tower and (B) beits foot. Let two balls meet at (C ) after time (t). Let AC =X,then B C =100 -x. . Taking vertical downwawnword motion of the ball dropped from the top, we have u=0, a=9.8 ms^(-2), S=xt=t As, S=ut + 1/2 at^(2) :. x=0 +1/2 xx 9.8 xx t&(2) =4.9 t^(20 ...(i) Taking vertical upward motion of theball throun up from (B), we have u=25 ms^(-1), a=- 9.8 ms^(-2), S =(100 -x), t=t As, S=ut +1/2 at^(23) :. 100 -x 25t +1/2 (-9.8)t^(20 =25 t- 49 t^(2) ....(ii) Adding (i) and (ii) we have , 100 =25 t or t=4 s Putting this value in (i) we get, x=4.9 xx 16 =78.4 m Hence, the two balls will meet after 4 seconds at a distance 78.4 m below the top .

Form the top of a tower

100 m

in height a ball is dropped and at the same time another ball is projected vertically upwards from the ground with a velcoity of

25 m s − 1

. Find when and where the two balls will meet.

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Updated On: 27-06-2022

Text Solution Solution

Let (A) be the top of a tower and (B) beits foot. Let two balls meet at (C ) after time (t). Let

A C = X , t h e n B C = 100 − x .

.

Taking vertical downwawnword motion of the ball dropped from the top, we have

u = 0 , a = 9.8 m s − 2 , S = x t = t As, S = u t + 1 2 a t 2 :. x = 0 + 1 2 × 9.8 × t & ( 2 ) = 4.9 t 20 ...(i)

Taking vertical upward motion of theball throun up from (B), we have

u = 25 m s − 1 , a = − 9.8 m s − 2 , S = ( 100 − x ) , t = t As, S = u t + 1 2 a t 23 :. 100 − x 25 t + 1 2 ( − 9.8 ) t 20 = 25 t − 49 t 2 ....(ii)

Adding (i) and (ii) we have ,

100 = 25 t or t = 4 s

Putting this value in (i) we get,

x = 4.9 × 16 = 78.4 m

Hence, the two balls will meet after

4 sec o n d s at a distance 78.4 m below the top . Answer

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