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 s initial speed 200 m/s speed 12.50 m/ks height [25.50 m 0.50 kg Strategy mass 100 Neglect air resistance. Since the gravitational force is conservative, use conservation of mechanical energy. KE PE TE Solution release angle Let h be the height of release relative to ground level. Choose ground level as the zero of gravitational potential energy. Let vi be the initial speed of the ball, and let vr be the speed of the ball just before it hits the ground 15 - 45 30 According to conservation of mechanical energy, the mechanical energy of the ball at the point of release (n = h) must equal the mechanical energy at a point just above the ground, where yt = 0 15 --30 --45 start resetclear Display in a New Window Solve this for the ball's final speed vf. (Use any variable or symbol stated above along with the following as necessary: g.) Let h = 25.5 m and vi = 14.1 m/s. Find the speed of the ball just before it hits the ground m/s Find the ball's mechanical energy at the release point. Find the ball's mechanical energy just before it hits the ground

Explanation / Answer

given that

vi = 14.1 m/s

h = 25.5 m

vf = sqrt(vi2 +2*g*h)

(a)

vf = sqrt(vi2 +2*g*h)

vf = sqrt( (14.1)2 +2*9.8*25.5)

vf = sqrt(698.61)

v = 26.43 m/s

(b)

KEi = 1/2*m*v^2 = 0.5*m*(14.1)2 = 99.40 m j

PEi = m*g*h = m*9.8*25.5 = 249.9 m

mecahnical energy at the release point

KEi +PEi = 99.40m+249.9m = 349.3m J

(c)

mechanical energy just before it hits the ground

KEf+PEf = KEi+PEi (according to conservation of energy)

KEf + PEf = 349.3m J

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