Question

est (PHYSICS B - 1st hour)3 seconds13A Soldier threw a grenade set to detonate on impact at an angle of 32° at a target exactly 39maway from where he was standing. Assuming the grenade hit the target at the same exactheight he threw it from, how much time did he have to escape the blast zone before thegrenade went off?(3 Points)Enter your answer

Answer 1

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Firstly, let us analyze the data the problem has given us. This can be achieved through a drawing

Given the equations of a movement with acceleration, we can write them as the following

`S_y(t)=S_(0y)+V_0*sin(32°)*t+(at^2)/(2)`
`S_x(t)=S_(0x)+V_0*cos(32°)*t`

Given these equations and our data, we can replace some of its unknowns, and we're left with

`S_y(t)=V_0*0.53*t-5t^2`
`S_x(t)=V_0*0.848*t`

The first two equations were reduced to these ones by applying the data that the angle is 32°, and that the grenade hit at the same height. Now, we'll plug the information about the distance it reached. This will leave us with the following

`S_x(t_(impact))=V_0*0.848*t_(impact)=39`
`S_y(t_(impact))=V_0*0.53*t_(impact)-5*t_(impact)^2`

Then we're left with the following system

`\begin{gathered} 0.53V_0t_(impact)-5t_(impact)^2=0 \\ 0.848V_0t_(impact)=39 \end{gathered}`

On our first equation, we can divide it by t impact, as we know it is not 0

`\begin{gathered} 0.53V_0-5t_(impact)=0 \\ 0.848V_0t_(impact)=39 \end{gathered}`

We can then rearrange and get the following

`\begin{gathered} V_0=(5t_(impact))/(0.53) \\ 0.848V_0t_(impact)=39 \end{gathered}`

By replacing V0 on the lower equation, we get

`0.848*(5t_(impact))/(0.53)*t_(impact)=39`

And this gives us the following equation

`5t_(impact)^2=39*(0.53)/(0.848)`

Simplifying...

`t_(impact)^2=4.875`

Finally, applying the square root, we're left with

`t_(impact)\text{ = }2.21s`

So this is the time the Soldier had to escape