Answer:
time of collision is
t = 0.395 s

so they will collide at height of 5.63 m from ground
Explanation:
initial speed of the ball when it is dropped down is

similarly initial speed of the object which is projected by spring is given as

now relative velocity of object with respect to ball

now since we know that both are moving under gravity so their relative acceleration is ZERO and the relative distance between them is 6.4 m



Now the height attained by the object in the same time is given as



so they will collide at height of 5.63 m from ground
Answer:
Explanation:
a ) speed of passenger = circumference / time
= 2π R / Time
= 2 x 3.14 x 50 / 60
= 5.23 m /s
b )
centrifugal force = m v² /R
= (882 /9.8 ) x 5.23² / 50
= 77.47 N
Apparent weight at the highest point
real weight - centrifugal force
= 882 - 77.47
= 804.53 N
Apparent weight at the lowest point
real weight + centrifugal force
= 882 +77.47
= 959.47 N
c ) if the passenger’s apparent weight at the highest point were zero
centrifugal force = weight
mv² /R = mg
v² = gR
= 9.8 X 50
v = 22.13 m /s
d )
apparent weight
mg - mv² / R
= 882 - (882 / 9.8 )x 22.13²/50
= 882 + 882
= 1764 N
=
Answer:
The answer is for your question is :
Explanation:
True