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.
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Answered by
33
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
=================================
Hope It Helps You ☺
________________________________
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
=================================
Hope It Helps You ☺
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rohit710:
Yeah , According to the calculations the answer is coming 2.14 .
Answered by
3
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
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