Question

An attacker at the base of a castle wall 4 m high throws a rock straight up with speed 7.5 m/s from a height of 1.5 m

above the ground.
(a) Will the rock exceed the top of the wall?
(b) If so, what is its speed when it reaches the top of the wall?
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(c) If we don’t want the stone to exceed the top of the wall (we want the top of the wall to be the maximum height)
what the initial speed that the stone must have?

Answer 1

a) Yes, b)
`v\approx 2.686\,(m)/(s)`, c)
`v_(o) \approx 7.002\,(m)/(s)`

Explanation:

a) The maximum height reached by the rock is:

`\Delta y = (\left(0\,(m)/(s) \right)^(2)-\left(7.5\,(m)/(s) \right)^(2))/(2\cdot \left(-9.807\,(m)/(s^(2)) \right))`

`\Delta y = 2.868\,m`

`y = 1.5\,m + 2.868\,m`

`y = 4.368\,m`

Yes, the rock will exceed the top of the wall.

b) The speed when the rock reaches the top of the wall:

`v = \sqrt{\left(7.5\,(m)/(s) \right)^(2)+2\cdot \left(-9.807\,(m)/(s^(2)) \right)\cdot (4\,m-1.5\,m)}`

`v\approx 2.686\,(m)/(s)`

c) The initial speed required so that stone does not exceed the top of the wall is:

`v_(o) = \sqrt{\left(0\,(m)/(s) \right)^(2)-2\cdot \left(-9.807\,(m)/(s^(2)) \right)\cdot (4\,m-1.5\,m) }`

`v_(o) \approx 7.002\,(m)/(s)`