Respuesta :
Answer:
[tex]\begin{gathered} s_x=75m \\ v_y=14.7ms^{-1} \\ \theta=30.46^o \end{gathered}[/tex]Explanation: A baseball is thrown with some velocity, and it reaches the same height after 3.0s. the horizontal component of the velocity is 25m/s, we need to find (i) the range of the baseball (ii) the vertical component of the velocity, and (iii) the angle of the launch.
(i) Range of the baseball:
Since the range is the horizontal distance covered, and we know the horizontal velocity component and the total time of the flight, therefore the range can be calculated as follows:
[tex]\begin{gathered} \text{Range }\Rightarrow s_x=v_xt \\ \therefore\rightarrow \\ s_x=(25ms^{-1})\times_{}(3.0s)=75m \\ s_x=75m \end{gathered}[/tex](ii) The vertical component of the velocity:
For the vertical component of the velocity, we will resort to the equation of motion in the y-axis, from it we can solve for the vertical component of the velocity, this is done next:
[tex]y(t)=y_o+v_yt-\frac{1}{2}gt^2\Rightarrow(1)[/tex]Using (1) and the fact that we have to set it equal to zero because the baseball lands at that height, the vertical velocity can be determined as follows:
[tex]\begin{gathered} y(t)=y_o+v_yt-\frac{1}{2}gt^2\Rightarrow(1) \\ y_o=0 \\ t=3.0s \\ g=9.8ms^{-2} \\ \therefore\rightarrow \\ y_o+v_yt-\frac{1}{2}(9.8ms^{-2})t^2=0\Rightarrow v_y(3.0s)-\frac{1}{2}(9.8ms^{-2})\times(3.0s)^2=0 \\ \therefore\rightarrow \\ v_y(3.0s)=\frac{1}{2}(9.8ms^{-2})\times(3.0s)^2\Rightarrow v_y=\frac{1}{2}(9.8ms^{-2})(3.0s)=14.7ms^{-1} \\ v_y=14.7ms^{-1} \end{gathered}[/tex](iii) The angle of the launch:
The last step is to calculate the angle of the launch, this can be simply calculated as follows:
[tex]\begin{gathered} \theta=\tan ^{-1}(\frac{v_y}{v_x}) \\ \therefore\rightarrow \\ \theta=\tan ^{-1}(\frac{14.7ms^{-1}_{}}{25ms^{-1}})=30.45552817^{\circ} \\ \theta=30.46^o \end{gathered}[/tex]