Answer:
a) Yes, b) [tex]v\approx 2.686\,\frac{m}{s}[/tex], c) [tex]v_{o} \approx 7.002\,\frac{m}{s}[/tex]
Step-by-step explanation:
a) The maximum height reached by the rock is:
[tex]\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)}[/tex]
[tex]\Delta y = 2.868\,m[/tex]
[tex]y = 1.5\,m + 2.868\,m[/tex]
[tex]y = 4.368\,m[/tex]
Yes, the rock will exceed the top of the wall.
b) The speed when the rock reaches the top of the wall:
[tex]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)}[/tex]
[tex]v\approx 2.686\,\frac{m}{s}[/tex]
c) The initial speed required so that stone does not exceed the top of the wall is:
[tex]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) }[/tex]
[tex]v_{o} \approx 7.002\,\frac{m}{s}[/tex]
