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Juli2301 [7.4K]
3 years ago
15

A trouble-making youth is standing on a bridge, and wants to drop a water balloon on an unsuspecting passerby. A man is jogging

on a path below the bridge with a constant speed of 4.2 m/s. The bridge is 11.6 m above the ground. If the balloon is to land right at the jogger's feet, at what horizontal distance x from the bridge should he be when the youth drops the balloon?
Physics
1 answer:
DIA [1.3K]3 years ago
8 0

Answer:

Explanation:

Time needed for a balloon to drop from vertical rest a distance of 11.6 m

t = √(2h/g) = √(2(11.6)/9.8) = 1.538618

d = vt = 4.2(1.538618) = 6.462197...

d = 6.5 m

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4. The Coyote has an initial position vector of \vec r_0=(15.5\,\mathrm m)\,\vec\jmath.

4a. The Coyote has an initial velocity vector of \vec v_0=\left(3.5\,\frac{\mathrm m}{\mathrm s}\right)\,\vec\imath. His position at time t is given by the vector

\vec r=\vec r_0+\vec v_0t+\dfrac12\vec at^2

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The Coyote hits the ground when r_y=0:

15.5\,\mathrm m-\dfrac g2t^2=0\implies t=1.8\,\mathrm s

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r_x=\left(3.5\,\dfrac{\mathrm m}{\mathrm s}\right)(1.8\,\mathrm s)=6.3\,\mathrm m

5. The shell has initial position vector \vec r_0=(1.52\,\mathrm m)\,\vec\jmath, and we're told that after some time the bullet (now separated from the shell) has a position of \vec r=(3500\,\mathrm m)\,\vec\imath.

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3500\,\mathrm m=v_0(0.56\,\mathrm s)\implies v_0=6300\,\dfrac{\mathrm m}{\mathrm s}

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