A baseball thrown at an angle of 65.0 above the horizontal strikes a building 17.0 away at a point 5.00 above the point from which it is thrown. Ignore air resistance. Find the magnitude of the velocity of the baseball just before it strikes the building.

10.8 m/s,-39.7º below the horizontal

Explanation:

let V be the initial velocity

the horizontal component is ... V cos (65.0)

the vertical component is ... V sin(65.0)  

for horizontal ... d=V*t=Vcos (65.0)*t

                         17=V*1/2*t

                        34=V*t

for vertical ... h=-1/2*g*t^2+V sin(65.0)*t

5=-4.90m/s^2*t^2+(√3/2)Vt

substituting for V t ..

5.00 m - 17 √3 m = -4.9 m/s² t²

(5.00 - 17 √3) / -4.90 = t²

2.17 s = t

(a) 34.0 m = V * 2.17 s

             V = 16.6 m/s

use the horizontal and vertical equations of motion to find the flight time (t)  

then substitute back to find the initial velocity (V)  

let v be the final velocity

finding the components of the final velocity  

the horizontal component is the same as the initial velocity vh = V cos(65.0º) = 16.6 / 2 = 8.30 m/s

for the vertical component v_v = V sin(65.0º) - g*t

v_v = (16.6 * √3 / 2) - (9.80 * 2.17) = - 6.89 m/s

the horizontal component is unchanged from the initial velocity (no horizontal acceleration)  

use gravitational acceleration to find the vertical component  

use Pythagoras to find the magnitude of the final velocity  

(b) v = √(vh² + v_v²) = 10.8 m/s

use trig to find the angle  

the tangent of the angle is tan(Θ) = v_v / vh = -.830

             Θ = -39.7º

the ball is moving downward  

cant without the picture! ?