You shoot a plastic arrow horizontally, at an initial velocity of 10m/s. The mas

Question

You shoot a plastic arrow horizontally, at an initial velocity of 10m/s. The mass of the arrow is 0.10kg. The feathers cause an approximately constant drug force of 0.25N to the left (-i). Thus the acceleration due to air drag is -2.5m/s^2. Of course it's also accelerating in the vertical direction due to gravity, -9.8 m/s^2] The distance to the wall is 5m. The initial height of the arrow is 2m. Give the equation for the velocity as a function of time.(velocity is a vector) Give the equation for thr position as a function of time.(position in a vector) At what time will the arrow hit the wall? At what height will the arrow hit the wall? Obtain the initial potential energy. Obtain the initial total energy. Obtain the final kinetic energy, just before it hits the wall. WARNING: it is moving to the right and down here!!!! Obtain the final total energy. The ENERGY LOST is also called the WORK due to air drag. This is just the Final total energy minus the initial total energy (so (i) - (g)). Calculate this value. (Important: This will be a negative number). The WORK done by gravity is just the Final potential energy minus the initial potential energy (so(i) - (f)). Calculate this value. (This will be a positive number) The NETWORK is the sum of the work done by gravity plus the ENERGY LOST due to air drag (so (k) + (i)). Add these two values. Verify above by calculating the change IN KINETIC ENERGY: delta KE = KE_final - KE_initial. This should be the same as (m).

Explanation / Answer

a)

along the X-direction :

acceleration = ax = - 2.5

Vi = initial velocity = 10 m/s

Vfx = velocity at any time "t"

Vfx = Vi + at

Vfx = 10 - 2.5 t                         eq-1

along y-direction :

Vfy = 9.8 t

Velocity = (10 - 2.5t) i^ + 9.8 t j^

b)

along the X-direction :

acceleration = ax = - 2.5

Vi = initial velocity = 10 m/s

t = time

position is given as

x = Vi t + (0.5) at2

x = 10 t + (0.5) (- 2.5) t2

x = 10 t - (1.25) t2              eq-1

position in Y-direction is given as

Y = Vi t + (0.5) at2

Y = 0 t + (0.5) (- 9.8) t2

Y = - (4.9) t2              eq-2

Position = X i^ + Yj^ = (10 t - (1.25) t2 ) i^ - (4.9) t2

c)

along X-direction :

x = 10 t - (1.25) t2  

when arrow hits , x = 5

5 = 10 t - (1.25) t2  

t = 0.54 or 7.5 sec

d)

height dropped in time t = 0.54 s

Y = - (4.9) t2 = - 4.9 (0.54)2 = - 1.43 m

height = 2 + Y = 2 - 1.43 = 0.57

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