A bullet fired vertically at a velocity of 36m/s .after 45 the bullet hit the top of a bulid how height is a bulid?​

Explanation:

A

h

1

=h

2

using equation of motion :s=ut+

2

at

2

distance traveled in first second:

h

1

=45m

t

a

=

g

u

,where t

a

is time of ascent

t

a

=5s

so at t=2s last 3 sec of upward motion starts

height at t=2s is :

h=50×2−

2

10×4

h=80m

H

max

=

2g

u

2

Hmax=125m

distance traveled in last 3 sec h

2

=H−h

h

2

=125−80

h

2

=45m

therefore h

1

=h

2

weight of the box is given as

here this is placed on the inclined of angle 42 degree

now this weight will have two components one is along the incline other is perpendicular to it

so here we need to apply a force parallel to incline opposite to its component of weight

so here the component of weight is given as

here we have

so it requires 75 n force to move it upwards

option number c is the right answer.