Find the Ball\'s Final Speed The applet provides readouts for time, object speed

Question

Find the Ball's Final Speed The applet provides readouts for time, object speed and object height above the ground. Notice that the ball always hits the ground with the same final speed, regardless of the initial release angle. Show that this is so. time 0.00 Inilal speed 200- speed 1250 mvs height 25.50m 0.50 k 12.50 m/s Strategy 100 Neglect air resistance. Since the gravitational force is conservative, use conservation of mechanical energy TE KE PE release Let h be the height of release relative to ground level. Choose ground level as the zero of gravitational potential energy. Let v be the initial speed of the ball, and let v be the speed of the ball just before it hits the ground. cording to conservation of mechanical energy, the mechanical energy of the ball at h) must equal the mechanical energy at a point just above 0. the point of release ( e ground, where yt slart pauseres Chapter C. × % Chapter 07 -) ? Not secure www.webassign.net/web/ tudent Assignment Responses submmdep.1814363e startpausset)dear Display in a New Window Solve this for the bal's final speed v. (Use any variable or symbol stated above along with the following as necessary: g) Let h 25.5 m and v 11.5 m/s. Find the speed of the ball just before it hits the ground. Find the ball's mechanical energy at the release point. Find the ball's mechanical energy just before it hits the ground m/s KE+ PE Notice that the final speed does not depend on the initial release angle. Kinetic energy is not a vector. The inial kinetic energy of the ball dspends on the bai's nitial speed. velocity Remarks

Explanation / Answer

m*vi^2/2 + mgh = mvf^2/2

divide by (m/2)

vi^2 + 2gh = vf^2

vf = sqrt (vi^2 + 2gh)

Part B

vf = sqrt (11.5^2 + 2*9.81*25.5)

vf = 25.15 m/sec

Part C

KEi + PEi = mvi^2/2 + mgh

= m*(11.5^2/2 + 9.81*25.5)

= 316.28*m J

Part D

KEf + PEf = mvf^2/2 + 0

= m*25.15^2/2

= 316.26*m J

Provide the value of 'm' and I will calculate it further.

If m = 0.5 kg, then

KEi + PEi = 316.28*0.5 = 158.14 J

KEf + PEf = 316.26*0.5 = 158.13 J

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