You are outside in a big, level field, in the midst of which stands a narrow wal

Question

You are outside in a big, level field, in the midst of which stands a narrow wall, 33.2 m high. You are given a device that can launch a projectile, always with the same speed of 58.0 m/s.

a) first, you launch a porjectile from the top of the wall, at an angle of 46.4 degrees above the horizontal. How far from the base of the wall does the projectile land on the ground?(answer in m)

Next you launch another projectile form the top of the wall. After 5.5 seconds of flight, the projectile lands on the ground.

b) At what angle above the horizontal was the projectile launched this time? (answer in degrees)

c)How far from the base of the wall does the projectile land in this case? )answer in m)

Now you position the launcher on the ground, a distance of 41.2 m from the base of the wall, at an angle of 18.3 degrees above the horizontal.

d) how high above the ground does the projectile strike the wall?(answer in m)

e)Still positioned at the same distance 41.2 m from the base of the wall, what is the minimum launch angle with respect to the horizontal that will allow the prjectile to just make it over the top of the wall? ( answer in degrees)

f) what is the maximum distance from the bottom of the wall that you can position your launcher on the ground, such that its possible to make it over the wall? (answer in m)

Explanation / Answer

along horizantal

initial velocity vox = vo*costheta

ax = 0

from equaiton of motion

x = vox*T+ 0.5*ax*T^2

x = vo*costheta*T

T = x/(vo*costheta)......(1)

along vertical

voy = vo*sintheta

acceleration ay = -g = -9.8 m/s^2

displacement y = -30 m

from equation of motion

dy = voy*T + 0.5*ay*T^2

dy = (vo*sintheta*x)/(vo*costhetra) - (0.5*g*x^2)/(vo^2*(costheta)^2)

(yf-yi) = x*tanthheta - ((0.5*g*x^2)/(vo^2*(costheta)^2))

(a)

(0-33.2) = x*tan46.4 - (0.5*9.8*x^2/(58^2*(cos46.4)^2))

x = 372 m <<<<<<---------answer

________

(b)

(yf-yi) = voy*T + 0.5*ay*T^2

(0-33.2) = (58*sintheta*5.5) - (0.5*9.8*5.5^2)

theta = 21.13 degrees

____________

(c)

X = vox*T = 58*cos21.13*5.5 = 297.6 m

___________

(d)

along vertical

v^2 - voy^2 = 2*ay*(yf-yi)

v = 0

0^2 - (58*sin18.3)^2 = -2*9.8*(yf-0)

yf = 16.9 m <<--------------answer

(e)

(yf-yi) = x*tanthheta - ((0.5*g*x^2)/(vo^2*(costheta)^2))

(33.2-0) = 41.2*tantheta - ((0.5*9.8*41.2^2)/(58^2*(costheta)^2))

theta = 42.5 degrees

___________

f)

along vertical

(yf-yi) = vo*sintheta*T - (0.5*g*T2)

(33.2-0) = (58*sin42.5*T)-(0.5*9.8*T^2)

T = 0.96 s   and T = 7.03

X = vox*T = 58*cos42.5*7.03 = 300.62 m   <<<<<----------answer

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