Question

11. From a height, two balls are thrown with equal
speed, one vertically upwards and other vertically
downwards. They take time t1, and t2 respectively
to reach the ground. If a third ball is dropped from
same height, then time taken to reach the ground
will be
(1) t1+t2/2
(2) √t1t2
(3) 2t1t2/t1+t2​

Answer 1

Answer:

T = √T₁T₂

Step-by-step explanation:

From a height, two balls are thrown with equal  speed, one vertically upwards and other vertically  downwards. They take time t1, and t2 respectively  to reach the ground. A third ball is dropped from  same height,

Let say Ball is thrown with Velocity V  from height H

S = ut + (1/2)at²

for dropped case

S = H  ,  u = V   a = g  t = T₁

H = VT₁  + (1/2)gT₁²    eq  1

Ball thrown upward

Take time = Time to go up & time to come at same point + T₁

=> T₂ = 2 * Time to go up + T₁   ( as time to go up & come back to same point is equal)

=> Time to go up = (T₂ - T₁ )/2

at top speed = 0

=> 0 = V - g (T₂ - T₁ )/2

=> V = g (T₂ - T₁ )/2

Putting this in eq  1

H =  g (T₂ - T₁ )T₁/2  + (1/2)gT₁²

if ball is dropped then time taken

H = (1/2)gT²  ( as initial velocity = 0)

=>  (1/2)gT²  =  g (T₂ - T₁ )T₁/2  + (1/2)gT₁²

=> T² =  (T₂ - T₁ )T₁  + T₁²

=> T² = T₂T₁ - T₁² + T₁²

=> T² = T₂T₁

=> T = √T₂T₁

=> T = √T₁T₂

option 2

Answer 2

(3) 12 kim From a height two balls are thrown with equal downwards.They take time t_(1) and t_(2) respectively to reach the ground.If a third ball is dropped from to reach the ground.If a third ball is dropped from same height then time taken to reach the ground will be (1) (t_(1)+t_(2))/(2) (2) sqrt(t_(1)t_(2)) (3) (2t_(1)t_(2))/(t_(1)+t_(2)) (4) (sqrt(t_(1)^(2)+t_(2)^(2)))/(2)