Question

1. A ball is thrown vertically upwards with a speed of 10 m/s from the top of a tower 200 m height and another is thrown vertically downwards with the same speed simultaneously. The time difference between them on reaching the ground is in second) (g= 10 m/s2) 1)12 2) 6 3)2 4) 1


I want explaination don't post single answer​

Answer 1

Answer:

For simpler understanding let’s consider the gravity to be 10m/s^2

and the time the balls are thrown to be t-0

The total time taken by Ball-1 from t-0 to the moment it reaches ground be t-1 and Ball-2 be t-2

We need (t-1) - (t-2)

t-1

Upward phase, u = 10m/s a= -10m/s^2 v=0, time = v-u/a = -10/-10 = 1 second

Height attained during intial phase = ut + 1/2gt^2 = 10*1 + 1/2(-10)(1)^2 = 5m

Downward Phase, u = 0 height = 205 m,

==> 205m = 0 + 1/2(10)(t^2)

t= Sqrt(41)

t-1 = 6.4031+1 = 7.4031

t-2

Height traveled = 200 m, u = 10m/s

==> 200 m = 10t + 1/2(10)(t^2)

5t^2+10t-200 = 0

t^2 + 2t - 40 = 0

t= [-2(+-)Sqrt(4–(4)(1)(-40))]/2 = [-1(+-)(1/2(Sqrt(164))] = -1 (+-) 12.8062/2 = 5.4031

Difference = 7.4031–5.4031 = 2 Seconds.

Answer 2

Answer:

naa peru yashaswini

nenu class 11 from hyderabad

nenu narayana college mpc

meru....?