Question

Question 23: 3 M
A hiker stands on the edge of a cliff 490 m above the ground and throws a
stone of mass 2 kg (at t=0) ,horizontally with an initial speed of 15 m 8-1.
Neglecting air resistance, find (a) the linear momentum of the stone just
before hitting the ground (in vector form). ,(b) average velocity of the
stone from t=0 to t=5 fecond. and (c) average acceleration of the stone
from t= 1 sec to t=9 sec. (Take g = 9.8 m 8-2 ).
Normal
BI U S 99
iii
li
M
Ix fx X₂ X²
Hi

Answer 1

motion of stone may be considered as the 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/s2

Let h be the height of the cliff above ground

Let uv be the vertical component of velocity of projection of the stone If the stone hits the ground after 1 second of projection the h=uvt+21gt2

the stone is thrown horizontally vertical component 

∴h=0+21gt2⇒t=92h

This gives time for stone to reach ground

t==2h99.82×490=9.8980=100=10Δ

t=10Δ

Let V4 be the vertical component of velocity of stone when it hots the ground then

vy2±uy2+2gh

vy2=(0)2+2×+9.8×490