Two towers A and B, each of height 20m are situated a distance 200m apart. A body thrown horizontally from the top of the tower A with a velocity `20 ms^(-1)` towards the tower B hits the ground at point P and another body thrown horizontally from the top of tower B with a velocity `30ms^(-1)` towards the tower A hits the ground at point Q. if a car starting from rest from P reaches Q in 10 seconds, then the acceleration of the car is (acceleration due to gravity `=10 ms^(-2)`)

Options

A. `1 ms^(-2)`
B. `2ms^(-2)`
C. `3ms^(-2)`
D. `4ms^(-2)`

Answer

B

Explanation

Given height of both towers is same,
`h_(1) = h_(2) = h`

Time of flight, `t = sqrt((2h)/(g))` will be same
`t = sqrt((2 xx 20)/(10)) = sqrt4 = 2s`
`rArr` Displacement in horizontal direction from tower A to point `P = u_(A) t`
`=20 xx 2 = 40m`
`rArr` Displacement in horizontal direction from tower B to point `Q = u_(B) t`
`=30 xx 2 = 60m`
So, distance between point P and Q
`=200 -(40 + 60)`
100m
Given, distance between P and Q is covered by car in 10s, so using
`s = ut + (1)/(2) "at"^(2)`
`100 = 0 xx 10 + (1)/(2) a (10)^(2)`
[ `:.` u = 0, as car starts from rest]
a =2
Acceleration, `a = 2m//s^(2)`