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, 49.6 m high. You are given a device that can launch a projectile, always with the same speed of 51.0 m/s.

Now you position the launcher on the ground, a distance of 69.3 m from the base of the wall, at an angle of 25.9 degrees above the horizontal.

a) How high above the ground does the projectile strike the wall? ————————————m

b)Still positioned at the same distance 69.3 m from the base of the wall, what is the minimum launch angle with respect to the horizontal that will allow the projectile to just make it over the top of the wall? ————————————degrees

Explanation / Answer

speed, u=51 m/sec

height , h=49.6 m

angle, theta=25.9 degrees

distnace, x=69.3 m

a)

use,

H=x*tan(theta)-1/2*g*x^2*(1/(u*cos(theta))^2)

H=69.3*tan(25.9)-1/2*9.8*69.3^2*(1/(51*cos(25.9))^2)

H=22.47 m

b)

use,

h=x*tan(theta)-1/2*g*x^2*(1/(u*cos(theta))^2)

49.6=69.3*tan(theta)-1/2*9.8*69.3^2*(1/(51*cos(25.9))^2)

====> theta=41.25 degrees

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