School and Work

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2 ways

Newton

d = 0.5at^2
30m = 0.5 * 9.8 m/s^2 * t^2

4.9 t^2 = 30
t^2 = 6.122
t = 2.47 s

v= at = 9.8 m/s^2 * 2.47 s = 24.25 m/s

Energy

PE = KE

MGH = 0.5MV^2 (cancel M)

GH = 0.5V^2
9.8 * 30 = 0.5 V^2
294 = 0.5V^2
V^2 = 294/.5 = 588
V = 24.25 m/s

No matter how you do it, the answer is always the same. (It would be really bad if it was not!)

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Home >All Categories > Science & Mathematics >Physics >Open Question Alana Member since:February 08, 2011Total points:116 (Level 1)A 20 g ball of clay traveling east at 3.6 m/s collides with a 30 g ball of clay traveling north at 2.0 m/s. What are the speed and the direction of the resulting 50 g ball of clay?Answer Question

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Home >All Categories > Science & Mathematics >Mathematics >Open Question green Member since:February 12, 2011Total points:79 (Level 1)a man walks along a straight path at speed 1.5m/s. A searchlight is located on the ground 6m from the path and is kept focus on the man. At what rate is the searchlight rotating when the man is 8m from the point on the path closest to the searchlight?Answer Question

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A disk-shaped merry-go-round of radius 2.23m and mass 165 kg rotates freely with an angular speed of 0.701 rev/s. A 63.4 kg person running tangential to the rim of the merry-go-round at 3.51m/s jumps onto its rim and holds on. Before jumping on the merry-go-round, the person was moving in the same direction as the merry-go-round's rim.

wf= ?rad/s

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A shot putter puts a shot through a horizontal distance 12.93 m. At the instant when the athlete releases the short, the elevation of the shot is 1.66 m above the ground level. The angle of release measured from the horizontal is 30°.

Determine the speed, in m s-1, at which the athlete releases the shot. Give your answer to 3 decimal places.

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