Finding missing quantities of a ball thrown at a building?

Question

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)

Answer 1

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 $V_i$.

Answer 2

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

With:

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

Now modify the formula