An archer fires an arrow horizontally at a target which is 25 m away, and at the same elevation. The initial speed of the arrow is 35 m/s.

a) How long is the arrow in the air?

b) By how much does it miss the center of the target?

c) The archer decides to correct his aim, and fires another arrow at the same speed, but at an angle of 3.5 upwards from the horizontal. By how much does this arrow miss the center of the target?

Solution:

a) no acceleration in the x-direction, so:

        x = x₀ + v_0xt

        25 m = 0 + (35 m/s) t

        t = 25m/(35 m/s)   =  0.714 s

b) arrow misses the target because it falls in y as it travels in x;

        y = y₀ + v_0yt - (1/2)g t²

where y₀=0   and v_0y= 0   so         y = - (1/2)g t²   =  - (1/2)(9.8 m/s²(0.714 s)²   =   -2.5 m

c) Now we no longer have v_0y=0;   instead         v_0y = v₀ sin(3.5) = 2.14 m/s

        v_0x = v₀ cos(3.5) = 34.93 m/s

        x = v_0xt

but otherwise we have the same thing as in a) and b);

        t = 25m/(34.93 m/s) = 0.716 s

        y = y₀ + v_0yt - (1/2)g t²   =   (2.14 m/s)t - (1/2)g t²

        = (2.14 m/s)(0.716 s)- (1/2)(9.8 m/s²(0.716 s)²   =   -0.98 m

so it still misses the center of the target; the archer must aim higher (or shoot with a larger initial speed!)

Test 1
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