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chubhunter [2.5K]
3 years ago
8

An archer fires and arrow while standing atop a 5.15 m tall wall. The arrow is fired at an angle of 55 degrees and has a launch

velocity of 52.4 m/s. (This is the actual launch velocity for English Longbows!) a. What is the time of flight for the arrow?
b. How far from the wall does the arrow strike the ground?
Physics
1 answer:
il63 [147K]3 years ago
8 0

Answer:

Explanation:

Given

height of wall=5.15 m

angle of launch(\theta )=55^{\circ}

Launch velocity(u)=52.4 m/s

Time of flight will be sum of time of flight of projectile+time to cover 5.15 m

Time of flight of arrow=\frac{2usin\theta }{g}

t=\frac{2\times 52.4\times sin55}{9.81}=8.76 s

Now time require to cover 5.15 m

Here at the time of zero vertical displacement of arrow i.e. when arrow is at the same height as of building then its vertical velocity will change its sign compared to initial vertical velocity.

v_y=-52.4sin55 at zero vertical displacement

Thus time required will be t_2

5.15=52.4sin55\times t+\frac{gt^2}{2}

4.9t^2+42.92t-5.15=0

t=0.118 s i.e. t_2=0.118 s

total time =t_1+t_2=8.76+0.118=8.878 s

(b)Horizontal distance=Range of arrow(R_1) + horizontal distance in 0.118 s(R_2)

R_1=\frac{u^2sin2\theta }{g}=\frac{52.4^2\times sin110}{9.8}=263.28 m

R_2=52.4cos55 \times 0.118=3.546 m

R=R_1+R_2=263.28+3.546=266.82 m

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3. A model rocket takes 0.05 seconds to speed up from rest to its maximum velocity of 80 m/s.
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Therefore we have:

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The velocity of the fish relative to the water when it hits the water surface is equal to the resultant velocity between the fish and the water when it hits it.

The fish drops on the water surface vertically with a vertical velocity v. Nothing was said about the velocity of the water, hence we can safely assume that the velocity if the water in the lake is zero, meaning that it is still. Therefore the relative velocity becomes equal to the velocity v with which the fish strikes the water surface.

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v = u + gt....................(1)

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It should also be noted that the horizontal and vertical components of the motion are independent of each other, hence we take u = 0 as the fish falls vertically.

To obtain t, we use the second equation of motion as stated;

h=ut+gt^2/2.................(2)

Given; h = 6.10m.

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h=\frac{1}{2}gt^2............(3)

substituting;

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Putting this value of t in equation (1) we obtain the following;

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