1(A) Height of the building is given 23m
Velocity with which ball is projected upward, u=20m/s
acceleration due to gravity is given, g=9.8m/s2
At maximum height, velocity of ball will be zero. i.e. v=0
using equation, v2=u2−2gh
0=(20)2−2∗9.8∗h
h=20.4m above the building.
For height above the ground,
H=20.4+23=43.4m
(B) time taken in upward motion, tu
v=u−gtu
putting values,
0=20−9.8∗tu⟹tu=9.820 s
Time taken in downward motion,
using equation, H=21gtd2 as initial velocity for downward motion will be zero.
then
td=g2H=9.82∗43.4
total time for motion, t=td+tu=9.820+9.82∗43.4=5.012s
2 Given equation is
(3x+2)+(3y−3)dxdy=0
(3x+2)dx+(3y−3)dy=0
comparing equation with Mdx+Ndy=0
M=3x+2 and N=3y−3
For equation to be exact, it must follow,
∂y∂M=∂x∂N
∂y∂M=0 and ∂x∂N=0
Hence equation is exact.
Integrating both sides,
∫(3x+2)dx+∫(3y−3)dy=C where C is constant.
23x2+2x+23y2−3y=C
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