Re: Hero velocity problem

let $\displaystyle t$ = time (in sec) the grenade is a projectile

position of the enemy car in meters ...

$\displaystyle x = 14.5 + \frac{515}{18} \cdot t$

position of the projectile ...

$\displaystyle x = \left(v_x + \frac{215}{9}\right)t$

... where $\displaystyle v_x$ is the x-component of grenade's initial velocity

since the idea is for the projectile and enemy car to be at the same place at the same time ...

$\displaystyle \left(v_x + \frac{215}{9}\right)t = 14.5 + \frac{515}{18} \cdot t$

in the y-direction, the projectile's total displacement is zero ...

$\displaystyle 0 = v_y t - \frac{1}{2}gt^2$

... where $\displaystyle v_y$ is the y-component of the grenade's initial velocity.

solving for $\displaystyle t$ ...

$\displaystyle t = \frac{2v_y}{g}$

since the launch angle is $\displaystyle 45^\circ$ , $\displaystyle v_y = v_ x$ , so $\displaystyle t = \frac{2v_x}{g}$.

... substitute for $\displaystyle t$ in the equal position equations and solve for $\displaystyle v_x$

Re: Hero velocity problem

If vy=vx wouldn't vx be 0 then?

Re: Hero velocity problem

Quote:

Originally Posted by

**Oldspice1212** If vy=vx wouldn't vx be 0 then?

how did you arrive at that conclusion?

Re: Hero velocity problem

I have no clue what I'm doing :(. How do you get the relative velocity?!

Re: Hero velocity problem

Quote:

If the hero throws the grenade so its initial velocity relative to him is at an angle of 45 degrees above the horizontal, **what should the magnitude of the initial velocity be?**

I've set up the necessary equations for you to solve for the magnitude of the grenade's initial velocity. If you cannot follow the proposed solution, then I advise you have a face-to-face consultation with your instructor.

Re: Hero velocity problem

Re: Hero velocity problem

Thought I had it, guess not, I got 54.7 km/h?