Question

A baseball thrown at an angle of 60 above the horizontal strikes a building

Answer 1

a) Let initial velocity be U

17= Ucos65 * T = 0.4226UT

T= 17/(0.4226U)= 40.2/U

H= Usin65 *T + 1/2 gT^2

6 = U*0.906 T + 1/2 * (-9.8) * T^2

6 = T ( 0.906U - 4.9T)

6= (40.2/U) (0.906U - 4.9*40.2/U)

6= 36.42 - 7918.6/U^2

7918/U^2 = 30.42

U^2 = 7918/30.42

U = 16.1 m/s

similarly apply sin and cosine rule to find the final velocity.

hope it helps you