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3 years ago

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An artillery shell is launched on a flat, horizontal field at an angle of α = 40.8° with respect to the horizontal and with an i

nitial speed of v0 = 346 m/s. What is the horizontal distance covered by the shell after 5.03 s of flight? Submit Answer Tries 0/12 What is the height of the shell at this moment?

1 answer:

Lina20 [59]3 years ago

8 0

Answer:

1317.4 m

Explanation:

We are given that

Angle= $\alpha=40.8^{\circ}$

Initial speed = $v_0=346m/s$

We have to find the horizontal distance covered by the shell after 5.03 s.

Horizontal component of initial speed= $v_{ox}=v_0cos\theta=346cos40.8=261.9m/s$

Vertical component of initial speed= $v_{oy}=346sin40.8=226.1m/s$

Time=t=5.03 s

Horizontal distance = $Horizontal\;velocity\times time$

Using the formula

Horizontal distance= $261.9\times 5.03$

Horizontal distance=1317.4 m

Hence, the horizontal distance covered by the shell=1317.4 m

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The motion of the rock is a projectile motion, consisting of two indipendent motions along the x- direction and the y-direction. In particular, the motion along the x- (horizontal) direction is a uniform motion with constant speed, while the motion along the y- (vertical) direction is an accelerated motion with constant acceleration $g=-9.8 m/s^2$ downward.

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For this part of the problem, we just need to consider the horizontal motion of the rock. The horizontal displacement of the rock at time t is given by

$x(t) = (v_0 cos \theta) t$

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For this part, we need to consider the vertical motion again.

We said that the vertical position of the rock at time t is

$y(t) = h+(v_0 sin\theta) t + \frac{1}{2}gt^2$

By substituting

y(t)=0

We find the time t at which the rock reaches the heigth y=0, so the time at which the rock reaches the ground:

$0=100+(30)(sin 53^{\circ})t+\frac{1}{2}(-9.8)t^2\\0=100+23.96t-4.9t^2$

which gives two solutions:

t = -2.69 s (negative, we discard it)

t = 7.58 s --> this is our solution

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The range of the rock can be simply calculated by calculating the horizontal distance travelled by the rock when it reaches the ground, so when

t = 7.58 s

Since the horizontal position of the rock is given by

$x(t) = (v_0 cos \theta) t$

Substituting

$v_0 = 30 m/s\\\theta=53^{\circ}$

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(e) (36.1 m, 128.3 m), (72.2 m, 117.4 m), (108.3 m, 67.4 m)

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$x(t) = (v_0 cos \theta) t$

$y(t) = h+(v_0 sin\theta) t + \frac{1}{2}gt^2$

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$y(2.0 s)= 100+(23.96)(2.0)-4.9(2.0)^2=128.3 m$

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$y(6.0 s)= 100+(23.96)(6.0)-4.9(6.0)^2=67.4 m$

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