What is the velocity by which a ball be projected vertically so that the distance covered by it in 5th second is twice the distance covered by it in 6th second(g=10 m/s^2)
A ball is projected vertically upwards, let the velocity at which it is projected be x m/s. The acceleration due to gravity is given as 10 m/s^2. The value of x has to be determined for which the distance covered by the ball in the 5th second is twice the distance covered in the 6th second.
The distance traveled by the ball in 4 seconds is given by (x*4 - (1/2)*10*4^2. The distance covered in 5 seconds is (x*5 - (1/2)*10*5^2. This gives the distance covered in the 5th second as (x*5 - (1/2)*10*5^2 - (x*4 - (1/2)*10*4^2) = x - 5*(25 - 16). Similarly, the distance covered in the 6th second is x - 5*(36 - 25). As the ball covers twice the distance in the 5th second than it does in the 6th, x - 5*(25 - 16) = 2*(x - 5*(36 - 25))
=> x - 45 = 2x - 110
=> x = 65 m/s
The ball is projected upwards at 65 m/s