Question

20) Two balls are porojected vertically
upwards in an interval of 2 second from
the ground.
The acceleration of one
body as seen by the other is
​

Answer 1

Answer:

75m

Explanation:

By Newton's 2 equation we get s = ut + 0.5 *a* (t^2)

s is the distance after time t , u is the initial velocity, a is acceleration.

When a ball is thrown up you can change the equation to +s = +ut + 0.5 *(- g) * (t^2)

where +s is the distance from the ground, +u is initial velocity in upward direction, (-g) is gravitational pull in -ve y axis.

so in order to find time t after the second ball is thrown

S1 = U1*(T+2) - 0.5*g*([T+2]^2) : FOR BALL 1

S1 = U2*(T) - 0.5*g*(T^2) : FOR bALL 2

Since when they collide S1 = S2 and UI = U2 (initial velocity)

We get :

U1*(T+2) - 0.5*g*([T+2]^2) = U2*(T) - 0.5*g*(T^2)

Lets say U1 = U2 = U

So

2U = 0.5 * g * { ([T+2]^2) - T^2 }

you get 2U / g = [2T + 2]

So T = [U/g] - 1.

Substitutiong U = 40 and g = 10m/s2 you get T = 3 seconds, if you take g = 9.8 it will be 3.04 sec

I'm taking g as 10 so t will be 3 sec.

Now just put the value in s = 120 - 5*9 = 75 m.