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3 years ago

8

Need help on this please

1 answer:

marin [14]3 years ago

3 0

The answer is D, the amount of energy stays the same.

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A student standing on the ground throws a ball straight up. The ball leaves the student's hand with a speed of 16.0 when the han

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There are mistakes in the question as the unit of speed and height is not mention here.The correct question is here

A student standing on the ground throws a ball straight up. The ball leaves the student's hand with a speed of 16.0m/s when the hand is 1.80m above the ground.How long is the ball in the air before it hits the ground? (The student moves her hand out of the way.)

Answer:

t=3.37s

Explanation:

Given Data

As we have taken hand at origin and positive upward

So given data are

$y_{i}=0m\\y_{f}=-1.80m\\v_{i}=16.0m/s\\a=g=9.8m/s^{2}$

To find

time taken by the ball before it hits the ground

Solution

By using the common kinematic equation

$y_{f}=y_{i}+v_{i}t+0.5at^{2}$

Put the given values and find for t

So

$-1.80=0+16.0t+(0.5*(-9.8)t^{2} )\\-1.80=16.0t-4.9t^{2}\\ 4.9t^{2}-16.0t-1.80=0$

Apply quadratic formula to solve for t

$t=\frac{-(-16.0)+\sqrt{(-16)^{2}+4(4.9)(-1.80)} }{2(4.9)}\\ t=3.37s$

4 0

3 years ago