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Nat2105 [25]
2 years ago
10

A very strong, but inept, shot putter puts the shot straight up vertically with an initial velocity of 11.0 m/s. how long does h

e have to get out of the way?
Physics
1 answer:
noname [10]2 years ago
7 0
The shot putter should get out of the way before the ball returns to the launch position.

Assume that the launch height is the reference height of zero.
u = 11.0 m/s, upward launch velocity.
g = 9.8 m/s², acceleration due to gravity.

The time when the ball is at the reference position (of zero) is given by
ut - (1/2)gt² = 0
11t - 0.5*9.8t² = 0
t(11 - 4.9t) = 0
t = 0 or t = 4.9/11 =  0.45 s

t = 0 corresponds to when the ball is launched.
t = 0.45 corresponds to when the ball returns to the launch position.

Answer: 0.45 s
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Answer:

T = 16.72 N

Explanation:

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At the bottom of the circle, the tension is directly upward, so these two forces, are opposite each other, and the difference between them is the centripetal force , which at this point, keeps the object swinging in a circle.

This is the point of the trajectory where T is maximum.

We can apply Newton's 2nd Law, choosing an axis vertical (y-axis) being the upward direction the positive one, as follows:

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The acceleration, at the bottom of the circle, is only normal (as there are no forces in the horizontal direction) , and is equal to the centripetal acceleration, as follows:

ac =  v² / r = ω²*r⇒ T- m*g = m*ω²*r

Replacing by the givens, we can solve for T as follows:

T = m* (ω²*r+g) = 0.4 kg*((8.00)² rad/sec²*0.5m)+9.8 m/s²) = 16.72 N

5 0
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Substituting numbers into the equation, we find

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