Question

From the top of a tower, two balls having identical masses are thrown in opposite directions, one at 3 m/s and the other at 4 m/s. Find the minimum distance at which their velocities are mutually perpendicular.
From Kinematics, 11th std.

Answer 1

Heya........!!!!

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See the pic in the attachment ↑↑

Nice question :)

=> According to the question balls are thrown in opposite direction so i have taken the appropriate direction of projection of balls , We cannot throw the balls in upward and downward direction because there will be no direction of velocities in perpendicular .

=> When balls are thrown in the direction according to figure i have drawn , there will be a point when there velocities will be mutually perpendicular to each other that is making an angle of 90° .

=> We know that velocities are Vector and when 2 vectors are at 90° there dot product is equal to 0 .


We have to write the velocity in Vector Form :-»

( applying the equation :- v = u + at )

•--» v1 => ( 3i^ - 9.8 t j^ )

•---» v2 => ( -4i^ - 9.8 t j^ )

Their dot product :-»

→ ( 3i^ -9.8 t j^ ) • ( -4i^ - 9.8 t j^ ) = 0

From this we can take out value of ' t '

Value of ' t ' comes => 15 / 49


Now we have to take out the distance ' s ' as shown in the figure , we will simply apply the formulae. ( distance = speed × time )

Here :

=> speed = 4 + 3 = 7
=> time = 15 / 49

•-――»» Distance => 7 × 15 / 49


♦ Distance => 2.14 m




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Hope It Helps You ☺

Answer 2

Hola mate❤✔

Answer:-

Let the first ball be thrown along the positive x-axis and the second one along the negative x-axis.

Velocity v1→ of the first ball after any time t is

v1→=9i^−10tj^

Similarly the velocity of the second ball after time t is

v2→=−4i^−10tj^

For the two velocities to be perpendicular, their dot product should be zero.

(9i^−10tj^).(−4i^−10tj^)=0⇒−36+100t2=0⇒t=0.6s

As the horizontal component of velocity remains constant, the total distance

x=0.6×9+0.6×4=7.8m