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

Answer 1

Answer:

a) Yes, b) $v\approx 2.686\,\frac{m}{s}$, c) $v_{o} \approx 7.002\,\frac{m}{s}$

Step-by-step explanation:

a) The maximum height reached by the rock is:

$\Delta y = \frac{\left(0\,\frac{m}{s} \right)^{2}-\left(7.5\,\frac{m}{s} \right)^{2}}{2\cdot \left(-9.807\,\frac{m}{s^{2}} \right)}$

$\Delta y = 2.868\,m$

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

$y = 4.368\,m$

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

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

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

$v\approx 2.686\,\frac{m}{s}$

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

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

$v_{o} \approx 7.002\,\frac{m}{s}$