Solve Car Chase Question: 110 km/h and 170 km/h

[anglum]

ok so can i move onto the next problem i have...

a ball is thrown at an angle of 50degrees at 17m/s ,,,, how long does it take to reach its maximum height??

 

[learningphysics]

ok so can i move onto the next problem i have...

a ball is thrown at an angle of 50degrees at 17m/s ,,,, how long does it take to reach its maximum height??

What is its vertical velocity at its maximum height?

 

[anglum]

im not sure what that is... these are my toughest problems

 

[learningphysics]

im not sure what that is... these are my toughest problems

At the max. height, the vertical velocity is 0. What is the initial vertical velocity?

 

[anglum]

the initial velocity is 17m/s but its at a 50 degree angle so what would its vertical velocity be

 

[learningphysics]

the initial velocity is 17m/s but its at a 50 degree angle so what would its vertical velocity be

17sin(50)

 

[anglum]

ok so the vertical velocity is 13.022m/s? so then i can plug that into the

Vf= Vi - AT

and solve for T and that is the time to reach its vertical max?

 

[anglum]

this question has 2 more parts

this ball is shot by a person at a height of 2.626m and goes thru a hoop 3.048m high

what is the distance of the shot... and how long does it take to reach the hoop?

again its shot at 17m/s at angle of 50 degrees

 

[learningphysics]

What is the vertical displacement of the ball?

Use d = (vsin(theta))*t + (1/2)(-g)t^2 to find the time...

Then find the horizontal distance using vcos(theta)*t

 

[anglum]

what the heck is vsin? theta? -g?

the vertical displacement is .422 meters correct?

 

[learningphysics]

what the heck is vsin? theta? -g?

the vertical displacement is .422 meters correct?

Yes. Try to use the equation d = v1*t + (1/2)at^2 in the vertical direction... that's the equation I gave you v = 17. theta = 50. g = 9.8

 

[anglum]

but i don't know what d or t are so how do i solve for that?

 

[learningphysics]

but i don't know what d or t are so how do i solve for that?

d=0.422. solve for t.

 

[anglum]

but that won't be the answer for how long it takes the ball to travel the horizontal distance to the hoop... the hoop is x away from the person... how can i solve how long it takes to reach the hoop without knowing how far the hoop is from teh person?

 

[learningphysics]

but that won't be the answer for how long it takes the ball to travel the horizontal distance to the hoop...

it is the same time.

the hoop is x away from the person... how can i solve how long it takes to reach the hoop without knowing how far the hoop is from teh person?

It is the same time. 17cos(50)*t gives the horizontal distance.

 

[anglum]

ok so if i solve for t on that last equation u gave me...

.422= 17sin50 (t) = .5(-9.8)t squared

.422 = 13.022t + -4.9t squared

and then i get stuck in solving for t

 

[anglum]

as u can tell my algebra is strugglin tonite

 

[learningphysics]

ok so if i solve for t on that last equation u gave me...

.422= 17sin50 (t) = .5(-9.8)t squared

.422 = 13.022t + -4.9t squared

and then i get stuck in solving for t

use the quadratic equation.

 

[anglum]

god i feel so dumb right now ... how do i use the quadratic equation to solve that

this is bad ... and I am so sorry

 

[learningphysics]

as u can tell my algebra is strugglin tonite

no prob.

 

[anglum]

i can't even function as to the quadratic equation solving that for me...

 

[learningphysics]

god i feel so dumb right now ... how do i use the quadratic equation to solve that

this is bad ... and I am so sorry

that's ok. have a look at the quadratic formula here:

http://en.wikipedia.org/wiki/Quadratic_formula#Quadratic_formula

First arrange the equation in the correct form:

4.9t^2 -13.022t +0.422 = 0

now try to apply the quadratic formula using the formula in the link.

 

[anglum]

t = 2.62?

 

[learningphysics]

t = 2.62?

yes. looks right.

 

[anglum]

so then for the horizontal distance it is just 17 cos 50 (2.62)?

 

[learningphysics]

so then for the horizontal distance it is just 17 cos 50 (2.62)?

yes.

 

[anglum]

ok 2 more problems if u don't mind?

#1

a rotating cylinder 10 miles long and 4.9 miles in radius is in space...

acceleration of gravity is 9.8m/s squared

what angular speed must the cylinder have so that the centripetal acceleration at its surface equals the free fall acceleration on earth?

 

[anglum]

ok i got that problem on my own actually WOOO HOOO

now the 2nd problem if u don't mind

 

[anglum]

014 (part 1 of 2) 10 points
The speed of a point on a rotating turntable,
which is 0.131 m from the center, changes at a
constant rate from rest to 0.958 m/s in 1.95 s.
At t1 = 1.6 s, find the magnitude of the
tangential acceleration. Answer in units of
m/s2.
015 (part 2 of 2) 10 points
At t1 = 1.6 s, ¯find the magnitude of the total
acceleration of the point. Answer in units of
m/s2.

 

[anglum]

learning these are my last 2 problems ... sorry to keep bothering you

 

[learningphysics]

014 (part 1 of 2) 10 points
The speed of a point on a rotating turntable,
which is 0.131 m from the center, changes at a
constant rate from rest to 0.958 m/s in 1.95 s.
At t1 = 1.6 s, find the magnitude of the
tangential acceleration. Answer in units of
m/s2.

This is just the change in velocity divided by time. ie 0.958/1.95.

015 (part 2 of 2) 10 points
At t1 = 1.6 s, ¯find the magnitude of the total
acceleration of the point. Answer in units of
m/s2.

What is the centripetal acceleration?

 

[learningphysics]

ok 2 more problems if u don't mind?

#1

a rotating cylinder 10 miles long and 4.9 miles in radius is in space...

acceleration of gravity is 9.8m/s squared

what angular speed must the cylinder have so that the centripetal acceleration at its surface equals the free fall acceleration on earth?

use centripetal acceleration = v^2/r. At a radius r, v = rw.