Question

Two Particles are projected vertically upwards with speed 20m/s and 100m/s from a top of a tower if their time so flight are in ratio 1:3 find the height of the tower?

Answer 1

use second equation of motion separately for both..to find t1 and t2 in terms of h...using
g=-10,s=-h,u1=20,u2=100.
use the ratio condition to find the value of h..

Answer 2

calculated it??

Answer 3

k.can u show me where u got struck?
anyone else calculated it?

Answer 4

No not yet....

Answer 5

Let height of tower be h.
Therefore time for the first projectile =
u = +20
height = -h.Let time be t1.
-h = +20t1 - 5t1^2 (As g = -10)-----Eq(1)
For second particle.
u = +100
height = -h. Let time be t2.
- h = 100t2 - 5t2^2. ---------Eq(2)

Eq(1) = Eq(2) ( -h = -h)

100t2 - 5t2^2 = 20t1 - 5t1^2.

Now t2 = 3t1.

300t1 - 5(9t1^2) = 20t1 - 5t1^2

40t1^2 - 280t1 = 0
Since t1 is not 0.
=> 40t1 = 280
t1 = 7 seconds.
t2 = 21 seconds

Satisfying t1 = 7 seconds in eq(1)

-h = 20*7 - 5*49
=> h = 105m

Answer 6

Thxxx but how h1=h2

Answer 7

it is the final displacement of both balls..

Answer 8

@open_study1 the tower is the same for both the particles so h1 will be equal to h2