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. Ignore air resistance. Find the magnitude of the initial velocity of the baseball (the velocity with which the baseball is thrown).

Question
Asked Sep 18, 2019
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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. Ignore air resistance.

 

Find the magnitude of the initial velocity of the baseball (the velocity with which the baseball is thrown).

 

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Expert Answer

Step 1

The angle made by the ball with ground is 60o, and distance from starting point to the building is 16.0m and height reached by the ball is 6.0m.

Write the expression for horizontal component of velocity

d is the horizontal di stance
t is the time
v cos 60°
substitute 16.0m for d
16.0m
vcos 60°
32m
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d is the horizontal di stance t is the time v cos 60° substitute 16.0m for d 16.0m vcos 60° 32m

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Step 2

Write the expression for ...

substitute 6.0m for h
9.8 m/s2 for g
32m for t
h (sin 60xt
32m
6.0m sin 60° 32m
U15.2m/s
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substitute 6.0m for h 9.8 m/s2 for g 32m for t h (sin 60xt 32m 6.0m sin 60° 32m U15.2m/s

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