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A baseball thrown at an angle of $55.0°$ above the horizontal strikes a building $18.0\,m$ away at a point $5.00\,m$ above the point from which it is thrown. Ignore air resistance.

Find the magnitude of the initial velocity of the baseball ( the velocity at which it is thrown)

clearly all we know is that at horizontal displacement 18 the vertical displacement is 5 regardless of whether it hit the building or not ( the questions asks for the quantities right before it hits the building) so I cannot simply use 0 for the final velocity(unless that just coincidentally happened to be the highest point of the trajectory, but that isn't a pre condition)

$$\text{displacement}_{vertical} = 5 = V_{intial} \sin(55) t + \frac12(-g) t^2 $$

I'm missing $V_{intial}$ and $t$ , cant do anything)

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  • $\begingroup$ Welcome to Math.SE. Check out this post on formatting your equations with MathJax. $\endgroup$ – John Sep 30 '16 at 18:35
  • $\begingroup$ You also know (using your notation) $$\text{displacement}_{h} = 18 = V_{intial} \cos(55) t $$ $\endgroup$ – N74 Sep 30 '16 at 21:58
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Hint: You also know that

$$x(t) = x_0 + v_x(0)t + (1/2)a_xt^2.$$

What is the acceleration in the $x$ direction? What is $v_x(0)$ in terms of $V_i$ and your angle?

This will give you another equation for $t$ in terms of $V_i$.

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$$y=y_0+x.tan(\alpha)-\frac{gx^2}{2(vcos(\alpha))^2}$$

With:

  • $y$ = height of impact
  • $y_0$ = initial hight
  • $x$ = horizontal distance
  • $\alpha$ = throw angle
  • $g$ = gravity

Now modify the formula

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