Could someone explain how to do this? Thanks!! Constants Part A A baseball throw

Question

Could someone explain how to do this? Thanks!!

Constants Part A A baseball thrown at an angle of 60.0 above the horizontal strikes a building 16.0 m away at a point 6.00 m above the point from which it is thrown. gnore air resistance Find the magnitude of the initial velocity of the baseball (the velocity with which the baseball is thrown). u= m/s Submit uest Answer Part B Find the magnitude of the velocity of the baseball just before it strikes the building u= m/s Submit uest Answer Part C Find the direction of the velocity of the baseball just before it strikes the building ° below the horizontal Submit uest Answer

Explanation / Answer

(a).

x = Vo t cos

t = x/(Vo cos )

y = Vo t sin + ½ g t²

= Vo (x/(Vo cos )) sin + ½ g (x/(Vo cos ))²

y = x tan + ½ g (x/(Vo cos ))²

y = (16) tan 60° + ½ (-9.8) (16/(Vo cos 60°))² = 6

27.7128-4.9/(16/(Vo cos 60°))²=6

0.22567=(16/(Vo cos 60°))²

=16.84m/s

(b).

t = x/(Vo cos )

t = 16/(16.84 cos 60°)

t = 1.9 sec.

Vx = Vo cos

= 16.84cos 60°

Vx = 8.42 m/s (forward)

Vy = Vo sin + g t

= 16.84 sin 60° + (-9.8)(1.9)

Vy = -4.036 m/s (downward)

resultant of ball's velocity is:

Vr = (Vx² + Vy²)

= (8.42² + (-4.036)²)

Vr = 9.337 m/s



(c).

to determine direction of ball's velocity when hits the building,

tan = Vy/Vx

= -4.036/8.42

= -0.479

= -25.61°

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