Let h be the height of release relative to ground level. Choose ground level as

Question

Let h be the height of release relative to ground level. Choose ground level as the zero of gravitational potential energy. Let v_i, be the initial speed of the ball, and let v_f be the speed of the ball just before it hits the ground. According to conservation of mechanical energy, the mechanical energy of the ball at the point of release (y_i = h) must equal the mechanical energy at a point just above the ground, where y_f = 0. 1/2mvi^2 + mgy_l = 1/2 mvf^2 + mgy f 1/2 mvi^2 + mgh = 1/2 mvf^2 + 0 Solve this for the ball's final speed v_f. (Use any variable or symbol stated above along with the following as necessary: g.) v_f = Let h = 25.5 m and v_i = 11.6 m/s. Find the speed of the ball just before it hits the ground. v_f = m/s Find the ball's mechanical energy at the release point. KE_i + PE_i = J Find the ball's mechanical energy just before it hits the ground. KE_f + PE_f = J

Explanation / Answer

from conservation of energy

mgh+1/2mv0^2 = 1/2mv^2

v = sqrt(2gh+v0^2)

a) v = sqrt(2*9.8*25.5+11.6^2) = 25.19 m/s

b) initial mechanical energy = 1/2mv0^2+mgh = m(67.28+249.9) = 317.18 m J

c) final energy = 1/2mv^2 = 317.18 m J

HERE MASS M IS NOT GIVEN

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