Question

An archer shoots an arrow with an initial speed of 45 m/s, at an angle of 70° from the horizontal.

How much time does it take to reach the top of its trajectory? 4.13 s

What is its velocity (magnitude and direction) at the top of its trajectory?15.4m/s^2

What is its acceleration (magnitude and direction) at the top of its trajectory?9.8m/s^2

How high does it go?91.23m

A)Assume that the arrow hits the ground 2 meters below the level from which it was launched. What is its total time of flight?

B)How far does the arrow travel in the horizontal direction?

C)What is the arrow’s velocity (magnitude and direction) at the instant just before it hits the ground?

Answer #1

Time taken to reach the top = u sin a / g

= 45 sin 70 / 9.8

= 4.31 s

acceleration at the top will be 9.8 m/s2 direction will be downwards

maximum height = u^2 sin^2 a / 2g

= 45^2 (sin 70)^2 / 2*9.8 = 91.23 m

Y = Uyt - 0.5gt^2

-2 = 45sin70t - 0.5*9.8*t^2

4.9t^2 - 42.28t - 2 = 0

t = 42.28 + sqrt(42.28^2 + 4*4.9*2) / 9.8

= 8.67 s

B) X = Ux*t = 45 cos 70 * 8.67

= 133.5 m

C) Vx = Ux = 45 cos 70 = 15.4 m /s

Vy = Uy - gt = 45 sin70 - 9.8*8.67 = -42.67 m/s

V = sqrt(15.4^2 + 42.67^2) = 45.37 ^{o}

angle = tan^-1(42.67 / 15.4) = 70.15^{o} below the
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