Question

if three balls are drop from a same height H. first one is thrown vertically upward second one is thrown down and third one is just dropped. time are t1 t2 and t3 . then find the relation between there times and options are 1 t3=t2 +t1 2 option is t2 =t3/t1 third option is t3 =√t2+t1 and last option is t3=t1t2/t1+t2​

Answer 1

Answer:

Step-by-step explanation:

The horizontal one would have the faster velocity as it would have the horizontal and he vertical one at the same time. As both of their vertical components of initial velocity equal zero, they are equally fast vertically. However, as one is horizontally projected, it should be horizontally faster than the other one.

Answer 2

Answer:

case 1 : Ball 1 is thrown vertically upward from height h, with speed u m/s. time taken by Ball 1 to reach the ground is t1.

using formula, S = ut + 1/2 at²

here, S = -h , t = t1 and a = -g

so, -h = ut1 -1/2 gt1²

or, -2h = 2ut1 - gt1²

or, gt1² - 2ut1 - 2h = 0

or, t1 = {2u ± √(4u² + 8gh)}/2g

but, t1 ≠ {u - √(u² + 2gh)}/g because it is negative value.

so, t1 = {u + √(u² + 2gh)}/g .....(1)

case 2 : Ball 2 is thrown vertically downward from height h, with speed u m/s

time taken to reach the ground by Ball 2 is t2 .

so, using S = -h, u = -u and a = -g

then, -h =-ut2 - 1/2gt2²

or, -2h = -2ut2 - gt2²

or, gt2² + 2ut2 - 2h = 0

or, t2 = {-2u ± √(4u² + 8gh)}/2g

or, t2 = {-u ± √(u² + 2gh)}/g

but t2 ≠ {-u - √(u² + 2gh)}/g

so, t2 = {-u + √(u² + 2gh)}/g......(2)

case 3 : Ball 3 is dropped from height h,

Time taken to reach the ground by Ball 3 is t3

so, -h = 0 + 1/2 × -g × t3²

or, t3 = √{2h/g}......(3)

multiplying equation (1) and (2),

t1t2 = {u² + 2gh - u²}/g² = 2h/g

or, √t1t2 = √{2h/g} = t3 [ from equation (3) ]

hence, proved