Energy Talk!

ETC asked:


I need assistance in figuring out this problem: A 0.40 kg ball is thrown with a speed of 12 m/s at an angle of 33 degrees. a)What is its speed at its highest point? and b) how high does it go? (Use conservation of energy, and ignore air resistance.)

(For part a) I’m guessing that the conservation of energy equation doesn’t apply because there is no vertical velocity and only horizontal velocity, which would be 12 * cos 33 degrees)

Thanks!

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3 Responses to “Physics Ball Thrown and Conservation of Energy Help?”

1. chhavi_sikri Says:
   December 3rd, 2008 at 2:25 pm

   yes part a is right
   for part 2 use the equation
   height= u sintheta* u sintheta/ 2g

2. Pointy Says:
   December 3rd, 2008 at 3:44 pm

   However way a ball is thrown (whether at an angle or vertically upwards), the speed at its highest point is ALWAYS equal to zero.

   To determine its maximum height, the formula to use is

   Kinetic energy = Potential energy

   Kinetic energy = (1/2)mV^2

   where

   m = mass of the ball
   V = vertical component of the initial velocity = 12sin33

   Potential energy = mgh

   where

   m = already defined above
   g = acceleration due to gravity = 9.8 m/sec^2 (constant)
   h = maximum height attained by the ball

   Since KE =PE,

   (1/2)(m)(12sin33)^2 = m(9.8)(h)

   NOTE that since m appears on both sides of the equation, it will simply cancel out and solving for h

   h = 4.9 (12sin33)^2

   I trust that you can proceed with the required mathematical process from here.

3. eyeonthescreen Says:
   December 4th, 2008 at 6:33 am

   No no no…the conservation of energy still applies in a). Here’s how.

   Upon release the ball’s total energy TE(V) = 1/2 mV^2 = KE(V) the kinetic energy; where m = .4 kg and V = 12 mps. That energy is conserved at the highest point (H) where total energy TE(H) = mgH + 1/2 m (V cos(theta))^2 = PE(H) + KE(H), which is the potential energy and kinetic energy at the highest point. vx = V cos(theta), where theta = 33 degrees; this is the x component of the launch velocity. We assume vx = constant because no air drag is considered.

   From the conservation of energy TE(V) = 1/2 mV^2 = mgH + 1/2 m(V cos(theta))^2 = TE(H); so the conservation of energy does in fact apply even though there is no vertical velocity. The good news is that you need not invoke it for answering a) because vx = V cos(theta) is all you need. But you do need it for b).

   Solving the conservation of energy relationships we have V^2 = 2gH + V^2 cos(theta)^2; so that H = V^2(1 - cos(theta)^2)/2g = V^2 sin(theta)^2/2g H is measure above the launch point, if you wish the H to be measured from ground level you need to add in the height of the launch point above ground.