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Question

Two balls are projected vertically upward with the same speed of 35m/s in a 3s interval, find the height in meters at which both balls collide.


A

50

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B

30

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C

80

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D

100

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Solution

The correct option is A

50


Step 1: Given Data

The initial velocity of both the balls u1=u2=35m/s

Time taken by the first ball =t

Time taken by the second ball =t-3

Let the distance traveled by both the balls be S1 and S2.

Let the acceleration due to gravity be g=-10m/s2 since the balls are moving upwards.

Step 2: Formula Used

Newtown's second equation of motion is given as,

S=ut+12gt2

Step 3: Calculate the Time Period

Since balls collide, their height must be equal.

S1=S235t-12gt2=35t-3-12gt-3235t-12gt2=35t-35×3-12gt2-6t+9-12gt2=-35×3-12gt2+3gt-92g0=-35×3+3×10t-3×32×100=-35+10t-1510t=50t=5s

Step 4: Calculate the height

Therefore, the height can be calculated as,

h=35×5-12×10×52=175-125=50m

Hence, the correct answer is option (A).


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