Question

7. A bullet is fired from the ground making an angle of 30 degwith the horizontal with a speed of 900 m/s.How long it will take to reach its maximum height? (1 point)A. 064.19 sB. O 79.372 sC. 045.918 sD. O 12.353 sIE. O 58.456 s8. A bullet is fired from the ground making an angle of 20 degwith the horizontal with a speed of 1500 m/s.Calculate the maximum height reached? (1 point)A. O 20831.097 m

Answer 1

We will have the following:

First, we determine the middle point of the projectile travel:

$R=\frac{(900m/s)^2sin(2\ast30)}{9.8m/s^2}\Rightarrow R=71579.65072...m$

So, the middle point will be approximately half of that, then the time will be determined as follows:

$\begin{gathered} \frac{(900)^2sin(60)}{9.8\ast2}=900cos(30)t+\frac{1}{2}(0)t^2\Rightarrow900cos(30)t=\frac{(900)^2sin(60)}{9.8\ast2} \\ \\ \Rightarrow t=\frac{(900)^2sin(60)}{900cos(30)(9.8)(2)}\Rightarrow t=\frac{900sin(60)}{(9.8)(2)cos(30)} \\ \\ \Rightarrow t=\frac{2250}{49}s \end{gathered}$

So, the time it will take to reach the maximum height is approximately 45.918s.