f4 physical

(1) Assume g = 10m/s^2
s = 0t + (1/2)gt^2 = 5t^2
80 = 5t^2 => t = 4
Therefore object P will be in the air for 4 seconds
The displacement of Q = 5(t-1)^2
The separation between P and Q = 5t^2 - 5(t - 1)^2 = 5(2t - 1)
The maximum separation is when t = 4, i.e. 5(2*4-1) = 35m
(2) s = 0t + (1/2)gt^2 = 5t^2
Displacement in the first second = 5*1^2 = 5m
Displacement in the first 2 seconds = 5*2^2 = 20m
Displacement in the second second = 20 - 5 = 15m
The required ratio is 5 : 15 = 1 : 3

(1) When P is released and reaches on the ground, in that moment, the seperation between P and Q is maximum,

u = 0, a = 10, s = 80, t = ?
By s = ut + (1/2)(a)(t^2)
80 = (0)t + (1/2)(10)(t^2)
t = 4s

When t = 4, distance travelled by P is 80m
the distance travelled by Q = ?, t = 3 (released 1s after), a = 10, u = 0
By s = ut + (1/2)(a)(t^2)
s = 45m

The max seperation is 80 - 45 = 35m

(2)
The distance in first second = ?, u = 0, a= 10, t= 1
By s = ut + (1/2)(a)(t^2)
s = 5m

The distance in two seconds =?, u = 0 a = 10 , t = 2
By s = ut + (1/2)(a)(t^2)
s = 20m

The distance travelled in second second = 20 - 5 = 15m

The ratio = 5: 15 = 1:3