A hiker stands on the edge of a cliff 490m above the ground and throws a stone
horizontally with an intial speed of 15ms-1
. Neglecting air resistance, find the time taken by
the stone to reach the ground and the speed with which it hits the ground.
Answers
super position of the two independent sources given horizontal with constant velocity u=15 m/s
vertices motion with constant acceleration a=g=9.8 m/s
2
Let h be the height of the cliff above ground
Let u
v
be the vertical component of velocity of projection of the stone If the stone hits the ground after 1 second of projection the h=u
v
t+
2
1
gt
2
the stone is thrown horizontally vertical component
∴h=0+
2
1
gt
2
⇒t=
9
2h
This gives time for stone to reach ground
t=
=
2h
9
9.8
2×490
=
9.8
980
=
100
=10Δ
t=10Δ
Let V
4
be the vertical component of velocity of stone when it hots the ground then
v
y
2
±u
y
2
+2gh
v
y
2
=(0)
2
+2×+9.8×490
v
y
2
=9604⇒V
y
=
9604
=98 m/s
The horizontal component of velocity V
x
with which the hits teh ground it remain constant because there is no acceleration is horizontal direction
∴V
x
=u
x
=15 m/s
∴ final speed of stone
V=
V
x
2
+V
y
2
=
15
2
+98
2
=
9829
=99.14 m/s
∴ speed with which the stone hits the ground is 99.4 m/s
pls follow me and Mark me as branliest pls
And give me thanks