First slide

Rectilinear Motion

Question

A juggler keeps on moving four balls in air throwing the balls after regular intervals. When one ball leaves his hand (speed = 20 m s^-1), the position of other balls (height in
meter) will be (take g = 10 ms^-2)

Moderate

A

10,20,10
B

15,20,15
C

5,15,20
D

5,10,20

Solution

Time taken by the same ball to return to the hands of the juggler is $\frac{2 u}{g} = \frac{2 \times 20}{10} = 4 s$ .
So he is throwing the balls after 1s each. Let at some instant he throws ball number 4.
Before 1 s of throwing it, he throws ball 3. So the height of ball 3 is
$h_{3} = 20 \times 1 - \frac{1}{2} 10 (1)^{2} = 15 m$
Before 2 s, he throws ball 2. So the height of ball 2 is
$h_{2} = 20 \times 2 - \frac{1}{2} 10 (2)^{2} = 20 m$
Before 3 s, he throws ball 1. So the height of ball 1 is
$h_{1} = 20 \times 3 - \frac{1}{2} 10 (3)^{2} = 15 m$

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