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2 years ago

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ucsd reddit A watermelon is thrown down from the 7th floor of Urey Hall at UCSD. If the initial height of the watermelon is 21.0

meters, and it is thrown straight downward with an initial downward velocity of 3.00 m/s. How far will the watermelon have fallen from its starting height after 1.50 seconds?

1 answer:

maxonik [38]2 years ago

4 0

Answer:

15.525 feet

Step-by-step explanation:

GIVEN: A watermelon is thrown down from the $7th$ floor of Urey Hall at UCSD. If the initial height of the watermelon is $21.0\text{ meters}$ , and it is thrown straight downward with an initial downward velocity of $3.00\text{ m/s}$ .

TO FIND: How far will the watermelon have fallen from its starting height after $1.5$ seconds.

SOLUTION:

initial height of watermelon $=21\text{ meter}$

initial velocity of watermelon $=3\text{ m/s}$

acceleration due to gravity $=9.8\text{m/s}^2$

According to newton's law of motion

$S=ut+\frac{1}{2}at^2$

when $t=1.5$ seconds

$S=3\times1.5+\frac{1}{2}\times9.8\times1.5^2$

$S=4.5+11.025$

$S=15.525\text{ feet}$

Hence watermelon will have fallen 15.525 feet from its starting height after 1.5 seconds

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2 years ago

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