Question

A baseball thrown at an angle of 60.0 ? above the horizontal strikes a building 16.0 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 with which the baseball is thrown). Find the magnitude of the velocity of the baseball just before it strikes the building. Find the direction of the velocity of the baseball just before it strikes the building.

Answer 1

Let initial velocity is v

horizontal velocity=vcos(60)=0.5v

initial vertical velocity=vsin(60)=0.866v

time taken to cover 16 m horizontally=16/0.5v=32/v

in this time vertical distance travelled =5 m

5=(0.866v*32/v)-(0.5*9.8*(32/v)^2)

v=14.86 m/s

(b) time taken=32/v=2.152

vertical velocity at final point=0.866*14.86-(9.8*2.152)=-8.23

horizontal velocity=0.5v=7.43

total velocity=sqrt(8.23^2+7.43^2)=11.08 m/s

direction=tan^(-1)(-8.23/7.43)=-47.92 degree or 312.07 degree