Question

A small lead ball, attached to a 1.5-m rope, is being whirled in a circle that lies in the vertical plane. The ball is whirled at a constant rate of three revolutions per second and is released on the upward part of the circular motion when it is 0.75 m above the ground. The ball travels straight upward. In the absence of air resistance, to what maximum height above the ground does the ball rise?

Answer 1

Answer:

H=10.93 m

Explanation:

Given that

h'= 0.75 m

r= 1.5 m

ω = 3 rev/s

ω = 3 ₓ 2 π rad/s

ω = 6 π rad/s

Speed of the ball when it releases from the rope,V.

V= ω x r

V= 6 π  x 0.75

V=14.13 m/s

We know that acceleration due to gravity always in downward direction.

$V_f^2=V_o^2+2as$

The final speed will be zero

$0=14.123^2-2\times 9.81 \times h$

h = 10.18 m

So the height from the ground,H = h' + h

H= 10.18 + 0.75 m =10.93 m

H=10.93 m