Given:
the initial velocity of the ball is
[tex]u=28\text{ m/s}[/tex]height of the hill is
[tex]h=2.3\text{ m}[/tex]The angle at which the ball is projected
[tex]\theta=45^{\circ}[/tex]Required: time of flight range, maximum height and final velocity is to be calculated.
Explanation:
when a ball is struck at an angle
[tex]\theta=45^{\circ}[/tex]with initial velocity u then its velocity has two components that is given by
in y-direction, velocity is
[tex]\begin{gathered} u_y=u\sin\theta \\ u_y=28\text{ m/s }\sin45^{\circ} \\ u_y=19.79\text{ m/s} \end{gathered}[/tex]in x-direction velocity is ,
[tex]\begin{gathered} u_x=u\cos\theta \\ u_x=28\text{ m/s }\cos45^{\circ} \\ u_x=19.79\text{ m/s} \end{gathered}[/tex]maximum of height of the ball is given by
[tex]h=\frac{u_y^2}{2g}[/tex]Plugging all the values in the above formula we get,
[tex]\begin{gathered} h=\frac{(19.79\text{ m/s})^2}{2\times9.\text{8 m/s}^2} \\ h=19.98\text{ m} \end{gathered}[/tex]time of flight of the ball is given by
[tex]T=\frac{2u_y}{g}[/tex]plugging all the values in the above relation we get,
[tex]\begin{gathered} T=\frac{2\times19.79\text{ m/s}}{9.8\text{ m/s}^2} \\ T=4.03\text{ sec} \end{gathered}[/tex]Range of the ball is given by
[tex]R=\frac{u^2\times\sin2\theta}{g}[/tex]Plugging all the values in the above formula we get,
[tex]\begin{gathered} R=\frac{(28\text{ m/s})^2\times\sin90}{9.8\text{ m/s}^2} \\ R=80\text{ m} \end{gathered}[/tex]Thus, range is 80 m, and Time of flight is 4.03 sec and maximum height is 19.98 m.