Conservation of energy question

Question

A 100 g ball is thrown straight up in the air from a height of 1.5 m above the floor. The ball just reaches the ceiling 10 m above the floor. What was the initial speed of the ball?

 

What speed will the ball have the instant before it hits the floor?

Answer 1

No law of conservation of energy is required for this simple problem.

It needs simple formula for a particle thrown vertically up.

Half the problem has already been solved by R Phillip using this formula.

 

Initial velocity = 12.9 mt/sec

 

Second part-----

 

Here the particle is coming down( a free fall) from a height of 10 Mt

We need finding out its velocity when it touches the floor.

 

V^2=2*9.8*10

 

V=14 mt/sec

 

Answer 2

certainly you should understand conservation of energy and the formula for kinetic energy = (1/2)mv²

 

In this case, you could get the same answer simply from the formula (final V)² - (initial V)² = 2as

 

0 - V₀² = 2(-9.8m/s²)(8.5m)

 

-V₀² = -166.6

 

V₀ = √166.6

 

V₀ = 12.9 m/s

Answer 3

Janine,

 

You can solve this problem using Conservation of Energy and the Work - Energy theorem.  To apply to this problem first note that the height (or distance) that the ball has to reach after being thrown is (10 m - 1.5 m) = 8.5 m.

 

In the instant after the ball was thrown it will have an initial velocity v which means it will have initial kinetic energy

 

K_i = ½·mv².  It's initial potential energy U = mgh, if taken with respect to its initial height 1.5 m above the floor is

Ui = 0.  At the top of its trajectory 8.5 m above the height from which it was thrown, it velocity will drop to zero as it momentarily stops rising and, reversing its motion, begins to fall.  At his point it kinetic energy K = 0, since v= 0, but its potential energy U = mgh is now (0.1 kg)(9.8 m/s²)(8.5 m) = 8.33 J.  Since energy is conserved:

 

K_i + U_i = K(top) + U(top)

 

Since Ui and K(top) are both = 0

 

K_i = ½·mv² = U(top) = mgh ⇒ ½·mv = 8.33 J

 

Solving for v_i = sqrt[2(mgh)/m] = [2·gh]½  (notice m cancels, so we the mass of the ball has no bearing on its final velocity and we didn't really need to explicitly calculate the value of U above.

 

v_i = [2·9.8 m/s²·8.5 m]^½ = 12.9 m/s (Initial velocity)

 

For the speed when it hit's the floor simply reverse the process this time with initial velocity = 0 and height = 10 m, since it has to fall the full distance from the ceiling to the floor.  In the instance before it hits the floor h = 0, so

 

mgh = ½·mv_f² ⇒ v_f = [2·9.8 m/s²·10 m]^½ = 14 m/s².  (Final velocity)

Answer 4

By conservation of energy:

 

(1/2)mv² = mgh

 

Where m= the mass of the ball = 100 kg

v = the initial speed the ball is thrown

g = the acceleration due to gravity = 9.8 m/s²

h = height upward the ball travels = 10 m - 1.5 m = 8.5 m

 

Solving for v:

(1/2)mv² = mgh

v² = 2gh

v = √2gh

 

The ball travels up ward from 1.5m to 10m, a total height of h = 8.5m.

 

v = √2(9.8)(8.5)

v = √166.6 = 12.9 meters/sec              (initial speed of ball)

 

The object's speed when it hits the floor is

 

v = √2gh    Where in this case h = 10m because the ball falls that far

 

v = √2(9.8)(10)

v = √196 = 14 m/s                             (speed when it hits the floor)

Answer 5

hi Janine-- I am SOOO not a physics person, but I think you CAN'T determine the speed of the ball going up since you don't know the force the thrower was using. You DO know the force on the way down, which is just gravity, and gravity makes things fall at 32 feet in the first second, then adding another 32 feet per second with each next second. So if the ball were falling a long way, it would pick up speed.

 

BUT this ball is only falling 10 meters, which is about 390 inches, or 32.5 feet. So it will fall 32 feet in the first second, and when it's falling the remaining half foot during the 2nd second, it SHOULD be falling at 64 feet per second right before it hits.

 

I think...