Question

A body is projected vertically upwards with a velocity 'u'. it crosses a point in its journey at a height 'h' twice ,just after 1 and 7 second. the value of u in m/s is (g =10 m/s²​

Answer 1

Given:-

→A body is projected vertically upwards with a velocity 'u'

→It crosses a point in it's journey at a height 'h' twice, just after 1 and 7 second.

→Acceleration due to gravity(g) = 10m/s²

To find:-

→Value of 'u' (initial velocity of the body)

Solution:-

From the problem, we can understand that the body crosses a point in it's journey namely 'h' two times. Just after, 1 second and 7 second.

In these cases:-

•Acceleration due to gravity(g) will be -10m/s², as the body is going against gravity.

When t= 1 second:-

By using the 2nd equation of motion, we get:-

=> h = ut+1/2gt²

=> h = u(1)+1/2×(-10)×(1)²

=> h = u-5. -----(1)

When t= 7 second:-

=> h = ut+1/2gt²

=> h = u(7)+1/2×(-10)×(7)²

=> h = 7u-245. ----(2)

Now, by subtracting eq.1 from eq.2, we get:-

=> h-h = 7u-245-(u-5)

=> 0 = 7u-245-u+5

=> 0 = 6u-240

=> -6u = -240

=> u = -240/-6

=> u = 40m/s

Thus, the value of 'u' is 40m/s.