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Artemon [7]
3 years ago
5

Suppose you now take the ball and using a bat, pop it straight up into the air with a hang-time of 5.00 s (the hang time is how

long the baseball is in the air after it leaves the bat). Determine the height to which the ball rises before it reaches its peak.
Physics
1 answer:
dolphi86 [110]3 years ago
8 0

Answer:

<h3>30.66m</h3>

Explanation:

Using the equation of motion formula S = ut + \frac{1}{2}gt^2 where;

S is the height to which the ball rises

u is the initial velocity of the ball = 0m/s

a is the acceleration due to gravity = 9.81m/s²

t is the time taken by the ball in air = 5.0s

Note that the  time to rise to the peak is one-half the total hang-time = 5.0/2 = 2.5s

Substituting the given parameters into the formula above to get S:

S = ut + \frac{1}{2}gt^2\\\\S = 0(2.5)+ \frac{1}{2}(9.81)(2.5)^2\\\\S = 0+\frac{1}{2}(9.81)\times 6.25 \\\\S = \frac{61.3125}{2}\\ \\S = 30.65625m\\\\S \approx 30.66m

This means that the ball rises 30.66m before it reaches its peak.

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<u>Answer</u>:

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3. Hot air balloon - The air is warmed up by a heating element within the balloon, so the air jumps upwards. This induces the balloon to increase in size due to the inside trapping of the warm air. He removes a few of the warm air when the pilot commences to dive, and cold air takes place, enabling the parachute to drop.

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