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otez555
20 days ago
6

A friend tosses a baseball out of his second floor window with initial velocity of 4.3m/s(42degrees below the horizontal). The b

all starts from a height of 3.9m and you catch the ball 1.4m above the ground.
a) Calc the time the ball is in the air (ans. 0.48s)
b)Determine your horisontal distance from window (ans. 1.5 m)
c)Calc the speed of ball as you catch it (ans: 8.2m/s)

I dont get what 42 m below the horizontal is, can someone give me direction on how to do this?
Physics
2 answers:
Keith_Richards [2.2K]20 days ago
7 0
<span>Part b) Find your horizontal distance from the window (answer: 1.5 m)
Part c) Calculate the speed of the ball upon catching it (answer: 8.2 m/s)

I'm confused about what "42 degrees below the horizontal" means. Could someone provide guidance on how to approach this?</span>
Maru [2.3K]20 days ago
6 0

Conclusion:

Part a)

t = 0.48 s

Part b)

x = 1.5 m

Part c)

v = 8.23 m/s

Explanation:

The ball's velocity is given by

v = 4.3 m/s

Now, we consider its two components of velocity:

v_x = 4.3 cos42 = 3.19 m/s

v_y = 4.3 sin42 = 2.88 m/s

Part a)

Now looking at the vertical direction we have:

y = y_o + v_y t + \frac{1}{2}gt^2

1.4 = 3.9 - 2.88 t - 4.9 t^2

We thus have:

t = 0.48 s

Part b)

The distance traveled by the ball in the horizontal plane is expressed as:

x = v_x t

x = 3.19 \times 0.48

x = 1.5 m

Part c)

The speed in the x-direction remains constant:

So, we find:

v_x = 3.19 m/s

And the speed in the y direction is given by:

v_y = v_i + at

v_y = 2.88 + (9.8)(0.48)

v_y = 7.58 m/s

Thus, the final speed is:

v = \sqrt{v_x^2 + v_y^2}

v = 8.23 m/s

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