physic velocity problem help??????

Question

You toss a chocilate bar to your hiking companion located 8.6 m up a 39 degree slope.determine the initial velocity vector so that chocolate bar will reach your friend moving horizontal??

Answer 1

The bar needs to go up a distance h = d sin theta

By conservation of energy

initial KE = 1/2 m vy^2 = final PE = mgh

EDIT--In response to a question I got, I should have pointed out that I'm only using the VERTICAL component of KE. Of course the bar has horizontal KE which isn't changing throughout the problem.

so vy = sqrt (2gh) = sqrt (2gd sin theta) is the vertical component of the throw velocity.

The time that takes is given by h = 1/2gt^2, so
t = sqrt (2h/g)= sqrt (2 d sin theta / g)

It must cover a horizontal distance x = d cos theta in the time taken, t, so horizontal velocity will be given by:
vx = x/t = d cos theta / sqrt (2 d sin theta / g)

If you want the magnitude of v, use pythagorus:
v = sqrt (vx^2 + vy^2)
If you want the direction, use
angle = arctan (vy / vx)

Answer 2

Answer is 12.1 m/s at an angle 58.3 degrees with the horizontal.

Let the initial velocity = u at an angle x with the horizontal. Chocolate bar will be moving horizontally at the maximum height of its projectile motion.

Now, maximum height,
H = 8.6 sin 39 degrees = [u^2 * sin^2 (x) / 2g] ... ( 1 )

Time to reach maximum height, t = u * sin x / g

Horizontal distance covered during this time
= 8.6 cos 39 degres = u * cos x * t u^2 * sin x cos x / g ... ( 2 )

Dividing eqn. ( 1 ) by eqn. ( 2 ),

tan 39 degrees = ( 1/2 ) tan x

=> tan x = 2 tan 39 degrees = 1.62
=> x = 58.3 degrees

Plugging this in eqn. ( 2 ),

u^2 = 8.6 cos 39 degrees * g / [ sin 58.3 deg. cos 58.3 deg.]
=> u^2 = 65.5 / 0.447 = 146.53
=> u = 12.1 m/s

Answer 3

If the bar is moving horizontally with respect to a non-inclined surface, then it is at the top of its trajectory. Your friend is a distance d at an angle q from you so he is higher by:

h = d*sin(q)

Since the bar has no vertical speed at your friend, you can write its position in y (vertical direction) as:

y = 1/2 g t^2 = d*sin(q) --> t = sqrt(2*d*sin(q)/g)

Then veritcal speed at launch is:

v = v0y - g*t = 0 ---> v0y =gt = sqrt(2*d*g*sin(q))

Finally you friend is a horizontal distance, x away from you

x = d*cos(q)

The bar traves a time t = sqrt(2*d*sin(q)/g), and must cover the distance x in that time. Hence the horizontal speed is:

v0x = x/t = d*cos(q)/sqrt(2*d*sin(q)/g)

Finally, the angle the bar is thrown at withrespect to the horizontal is

w = arctan(v0y/v0x) = arctan(sqrt(2*d*g*sin(q))/(d*cos(q)/sqrt(2*d*sin(q)/g)) )

w = arctan(2*d*sin(q)/(d*cos(q))) = arctan(2*tan(q))

YOu can plug in the numbers g = 9.8 m/s^2

Answer 4

4) They strike the floor with a similar velocity. it particularly is least confusing to work out by utilising pondering potential: the two balls start up with a similar kinetic potential, so after dropping a similar volume of potential potential, they might desire to end with a similar kinetic potential. 5) Use the equation of action with acceleration with the aid of gravity: h = h0 + v0 t - g/2 t^2

Answer 5

Easy! Walk past your companion to a point which is 8.6 m BELOW you. You`ll find it easier that way.