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In Denver, children bring their old jack o lanterns to the top of a tower and compete for accuracy in hitting a target on the ground. suppose that the tower is 9.0m high and that the bullseye is a horizontal distance of 3.5m from the launch point.if the pumpkin is thrown horizontally what is the launch speed needed to hit the ground? (b) If the jack o lantern is given an initial horizontal speed of 3.3m/s, what are the direction and magnitude of its velocity (c) 0.75s later and (d) just before it lands.

In Denver Children Bring Their Old Jack O Lanterns To The Top Of A Tower And Compete For Accuracy In Hitting A Target On The Ground Suppose That The Tower Is 90 class=

Sagot :

a)

In order to calculate the required horizontal speed, first let's calculate the falling time, using the free-fall formula:

[tex]d=\frac{gt^2}{2}[/tex]

For d = 9 and g = 9.8, we have:

[tex]\begin{gathered} 9=4.9t^2 \\ t^2=\frac{9}{4.9} \\ t=1.355\text{ s} \end{gathered}[/tex]

Then, let's use this time in the following formula for horizontal distance, so we can calculate the horizontal speed:

[tex]\begin{gathered} \text{distance}=\text{speed}\cdot\text{time} \\ 3.5=\text{speed}\cdot1.355 \\ \text{speed}=\frac{3.5}{1.355}=2.58\text{ m/s} \end{gathered}[/tex]

b)

Let's calculate the vertical velocity after 0.75 seconds, using the formula:

[tex]\begin{gathered} V=V_0+a\cdot t \\ V=0+9.8\cdot0.75 \\ V=7.35\text{ m/s} \end{gathered}[/tex]

The magnitude can be calculated using the Pythagorean Theorem with the horizontal and vertical velocities:

[tex]\begin{gathered} V^2=7.35^2+3.3^2 \\ V^2=54.0225+10.89 \\ V^2=64.9125 \\ V=8.06\text{ m/s} \end{gathered}[/tex]

And the direction is given by the arc tangent of the vertical velocity divided by the horizontal velocity (for this, let's use a negative value of vertical velocity, since it points downwards)

[tex]\theta=\tan ^{-1}(-\frac{7.35}{3.3})=\tan ^{-1}(-2.2272)=-65.82\degree[/tex]

(d)

To find the vertical velocity, let's use Torricelli's equation:

[tex]\begin{gathered} V^2=V^2_0+2\cdot a\cdot d \\ V^2=0^2+2\cdot9.8\cdot9 \\ V^2=176.4 \\ V=13.28\text{ m/s} \end{gathered}[/tex]

Calculating the final speed magnitude and orientation, we have:

[tex]\begin{gathered} V^2=13.28^2+3.3^2 \\ V^2=187.2484 \\ V=13.68\text{ m/s} \\ \\ \theta=\tan ^{-1}(-\frac{13.68}{3.3})=\tan ^{-1}(-4.145)=-76.44\degree \end{gathered}[/tex]

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