A bullet is fired at an angle of 45 degrees to the horizontal with a velocity of 49 m/s calculate the time of flight horizontal range the maximum height attained by the bullet during its flight

Dear Student,

Please find below the solution to the asked query:

Given information is,

Initial velocity of the bullet is $U = 49 m/s$ and the bullet is fired at an angle $\theta ={45}^0$.

As we know the equation for Time of flight is,

$$\begin{gathered} T_f=\frac{2U \sin \theta }{g}=\frac{2\times 49\times \sin {45}^0}{9.8}=\frac{98}{9.8\sqrt{2}} \\ \Rightarrow T_f=5\sqrt{2} sec=7.07 sec \end{gathered}$$

The equation for the horizontal range is,

$$\begin{gathered} R=\frac{U^2 \sin 2\theta }{g}=\frac{{49}^2\times \sin 2\times {45}^0}{9.8}=\frac{49\times 49\times 10}{98}=5\times 49 \\ \Rightarrow R=245 m \end{gathered}$$

The equation for maximum height is,


$$\begin{gathered} h_{max}=\frac{U^2 {\sin }^2 \theta }{2g}=\frac{{49}^2\times {\sin }^2 {45}^0}{2\times 9.8}=\frac{49\times 49}{4\times 9.8}=\frac{490}{8} \\ \Rightarrow h_{max}=61.25 m \end{gathered}$$
 

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