Question

A baseball thrown at an angle of 55.0 above the horizontal strikes a building 17.0 m away at a point 8.00m above the point from which it is thrown. Ignore air resistance. A) Find the magnitude of the initial velocity of the baseball (the velocity with which the baseball is thrown). Express your answer in meters per second. B) Find the magnitude of the velocity of the baseball just before it strikes the building. Express your answer in meters per second. C) Find the direction of the velocity of the baseball just before it strikes the building. Express your answer in degrees.

Answer 1

- 17m (Rough diagram) o @ y = lusino) t-1 g +22 Equations of x= Cucoso) t motion for y and & axis. substituting It= 4 cool in o n ucoso We get (y= (hand) x - (autoto) x² G Standard I equation. a) How substitute the values. 4=8 ,X=17, q=9.8, 0=55° in equation ③ . 9,8 (17) = (Tan 55 alt 2u²cos²55 solving we get [u = 16.261 m/sec 8 u = 16,2

x-component does not change. Ux= ucoso = 9.3269 m/s Uy= usino-gt (or) uy?(u since) -2(g) (y) TVZ-u?=2as. uy = Jussinta-agy y = √ (16.261) ain'ss - 2 (9.8) (8) My = 4.541867935 urls / Magnitude of the baseball just before it hits the building = Jully² = 10.374 m/sec 4.541867935 e) tand- uy 9.3269 tano = 0.48696 Tor 25,964° ~ 26°)