A woman tries to throw a rock over a wall, releasing the rock at a height of 1.55m abovethe ground. If she throws the rock at 6 m/s, will it reach the top of the wall 3.75m aboveher?

Respuesta :

Answer:

[tex]H_{\max }-H_{\text{wall}}=-0.3635m\Rightarrow Negative[/tex]

Explanation: We need to find if the rock can reach the height of the wall, provided the initial velocity of 6m/s and it is thrown from the height of 1.55m above the ground., the equations used to solve this problem are as follows:

[tex]\begin{gathered} v(t)=v_o-gt\Rightarrow(1) \\ y(t)=y_o+v_ot-\frac{1}{2}gt^2\Rightarrow(2) \end{gathered}[/tex]

Plugging in the known values in the equation (1) and (2) we get the following results:

[tex]\begin{gathered} v(t)=6ms^{-1}-(9.8ms^{-2})t\Rightarrow(3) \\ y(t)=1.55m+(6ms^{-1})t-\frac{1}{2}(9.8ms^{-2})t^2\Rightarrow(4) \\ \end{gathered}[/tex]

Setting equation (3) equal to zero gives the time to reach the maximum height as follows:

[tex]\begin{gathered} v(t)=6ms^{-1}-(9.8ms^{-2})t=0 \\ t=\frac{(6ms^{-1})}{(9.8ms^{-2})}=0.61s \\ t=0.61s \end{gathered}[/tex]

Substituting the time t calculated above in equation (4) gives the maximum height, the steps for the calculation are shown as follows:

[tex]\begin{gathered} y(0.62)=1.55m+(6ms^{-1})(0.62s)-\frac{1}{2}(9.8ms^{-2})(0.62s)^2 \\ y(0.62)=1.55m+3.72m-1.88356m=3.3865m \\ y(0.62)=3.3865m \\ H_{\max }=3.3865m \\ H_{\max }-H_{\text{wall}}=3.3865m-3.75m=-0.3635m\Rightarrow Negative \\ \\ \end{gathered}[/tex]

Therefore, the ball can not reach the height of the wall!