Ayoungster throws a rock from a bridge into the river 50 m below. the rock has a speed of 15 m/s when it leaves the youngster’s hand. calculate the velocity of the rock when it strikes the water if it is thrown upward.

Firstly rock is thrown in upward direction with initial velocity u= 15m/s. As the rock reaches top most point, it's final velocity (v) becomes zero. Therefore height reached by rock during this interval can be obtained by kinematic equation since acceleration is constant. Therefore, v^2 - u ^2 = 2ah V=0, u= 15m/s, a=-9. 8m/s^2 ( minus since acceleration is in opposite direction or we can say rock experiences deacceleration). Putting these values in above equation we get h=11.48 m. Now rock falls from this point with initial velocity (u= 0). a= 9.8 (since rock will experience acceleration), h= 11.48 +50 = 61.48 m. Again using the same above given equation : v^2 - u ^2 =2ah. V^2 - 0 = 2 x 9.8 x 61.48

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