Solving Neutron Star Physics Problems: Speed, Acceleration, Height

[winlinux]

4)
A ball is shot from the ground into the air. At a height of 9.1m, its velocity is observed to be v =7.6 i + 6.1 j in meters per second. (a) To what maximum height does the ball rise? (b) What horizontal distance does the ball travel? What are (c) the magnitude and (d) the direction of the ball’s velocity just before it hits the ground?

5)
When a large star becomes a supernova, its core may be compressed so tightly that it becomes a neutron star, with a radius of about 20 km (about the size of the San Francisco area). If a neutron star rotates once every second, (a) what is the speed of a particle on the star’s equator and (b) what is the magnitude of the particle’s centripetal acceleration? (c) If the neutron star rotates faster, do the answers to (a) and (b) increase, decrease, or remain the same?

thanks!

 

[rgrig]

for 4: use v_2^2 - v_1^2 = 2 a \Delta x

 

[TenaliRaman]

4>
a> third kinematic equation
b> range equation or simple distance = speed * t would also do
c> u have to find the vertical velocity but think over it u may have already found it out during a and b maybe.

5>
a>v=rw
b>v^2/r
c>u decide!

-- AI

 

[winlinux]

4>
a> third kinematic equation
b> range equation or simple distance = speed * t would also do
c> u have to find the vertical velocity but think over it u may have already found it out during a and b maybe.

5>
a>v=rw
b>v^2/r
c>u decide!

-- AI

5a) v=40kpi (rad/s)??

4)
i already solve 4a)10.9605m
4b)s=ut+0.5at^2
-9.1=6.1t+0.5(-10)t^2
t=2.0906s
what is the horizontal speed? is it 7.1*2.0906=ans??

how to solve b) and c)??

 

[TenaliRaman]

Before going to your 5th let's finish ur 4th,
Assuming u r not from antarctica where u might be holding a map upside down relative to me, what would v =7.6 i + 6.1 j tell u ??

-- AI
P.S -> i assumed u know that horizontal velocity remains constant during a projectile motion

 

[winlinux]

Before going to your 5th let's finish ur 4th,
Assuming u r not from antarctica where u might be holding a map upside down relative to me, what would v =7.6 i + 6.1 j tell u ??

-- AI
P.S -> i assumed u know that horizontal velocity remains constant during a projectile motion

7.6 is the horizontal velocity
6.1 is the vertical velocity
so
4a) 0=6.1^2+2(-10)s
s=1.8605m
1.8605m+9.1m=10.9605m

4b)s=ut+0.5at^2
-9.1=6.1t+0.5(-10)t^2
t=2.0906s
what is the horizontal speed? is it 7.1*2.0906=15.8886m

is it correct??

how about c) and d) how to solve?

 

[TenaliRaman]

what is the horizontal speed? is it 7.1*2.0906=15.8886m

I think u mean horizontal distance ...
is that that t=2.0906 the overall time?

for c and d,
get the horizontal velocity when it hits the ground(say v_h)
get the vertical velocity when it hits the ground(say v_v)
magnitude = sqrt(v_h*v_h+v_v*v_v)
angle = arctan(v_v/v_h)

-- AI