A stone is thrown horizontally off a bridge 20m above a riverhas an initial velo

Question

A stone is thrown horizontally off a bridge 20m above a riverhas an initial velocity of 17m/s. Determine the time t the stone is in the air, therange of the stone and the magnitude of the velocity when the stonestrikes the water. I worked the equations as t = the square root of 2h/g = square root of 2 times 20g =square root of 40/-9.8 = 6.32/-9.8 = 0.6455s r = VxT = 17m/s times .644s = 10.94m magnitude of the velocity I was completely confusedwith. A stone is thrown horizontally off a bridge 20m above a riverhas an initial velocity of 17m/s. Determine the time t the stone is in the air, therange of the stone and the magnitude of the velocity when the stonestrikes the water. I worked the equations as t = the square root of 2h/g = square root of 2 times 20g =square root of 40/-9.8 = 6.32/-9.8 = 0.6455s r = VxT = 17m/s times .644s = 10.94m magnitude of the velocity I was completely confusedwith.

Explanation / Answer

The vertical distance y = 0.5gt2 Then we have                             t = 2y/g                               = 2*20/9.8                               =2.02 s The horizontal range of the stone travelled during this timeis                            x = vx * t                               = 17 * 2.02                               = 34.34 m The vertical velocity with which the stone hit the water                      vy = 2gy = 2*9.8*20 = 19.79 m/s The magnitude of velocity with which the stone hit the water= (vx2 +vy2)                                                = 26.08m/s

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