# Wally is shot from a cannon with a speed of 20 m/s at an angle of 30 degrees above horizontal. Luke drops from a platform (x,y) coordinates of (8,16).a)will the make contact?b)what is minimum...

Wally is shot from a cannon with a speed of 20 m/s at an angle of 30 degrees above horizontal. Luke drops from a platform (x,y) coordinates of (8,16).

a)will the make contact?

b)what is minimum distance separating wally and luke during their flight paths?

c)what time does minimum separation occur?

d)give coordinates of each person at that time.

justaguide | College Teacher | (Level 2) Distinguished Educator

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Wally is shot from a cannon with a speed of 20 m/s at an angle of 30 degrees above the horizontal. It is assumed that the canon was placed such that the initial point of launch had coordinates (0,0). Luke drops from a platform with (x,y) coordinates of (8,16).

The velocity at which Wally is shot can be divided into x and y components. The component along the x-axis is 20*cos 30 = `10*sqrt 3` and the component along the y-axis is 20*sin 30 = 10.

As Wally moves, the x-component is constant. The y-component decreases.

The time required by Wally to travel 8 units along the x-axis is `8/(10*sqrt 3)` . His height at this point of time is `(10*8)/(10*sqrt 3) - (1/2)*g*(8/(10*sqrt 3))^2` = `8/sqrt 3 - g*(32/300)` . Luke is dropped from the point (8, 16). Luke's height at time t is `16 - g*(32/300)` . As `8/sqrt 3 - g*(32/300) != 16 - g*(32/300)` , the two do not make contact.

The distance between Luke and Wally at time t is given by D = `sqrt((8 - 10*sqrt 3*t)^2 + (16 - (1/2)*g*t^2 - (10*t - (1/2)*g*t^2))^2)`

=> D = `sqrt((8 - 10*sqrt 3*t)^2 + (16 - 10*t)^2)`

`(dD)/(dt) = (100*t-20*sqrt(3)-40)/(sqrt(5)*sqrt(5*t^2+(-2*sqrt(3)-4)*t+4))`

Solving `(dD)/(dt) = 0` gives `(100*t-20*sqrt(3)-40) = 0`

=> t = `(20*sqrt 3 + 40)/100`

The value of `(d^2D)/(dt^2)` at `t = (sqrt 3 + 2)/5` is positive. This indicates a minimum value of D at `t = (sqrt 3 + 2)/5` .

At t = `(sqrt 3 + 2)/5` , the coordinates of Luke are `(8, (16 - (1/2)*9.8*((sqrt 3 + 2)/5)^2)` = (8, 13.27)

The x- coordinate of Wally is `2*sqrt 3*(sqrt 3 + 2) = 12.93` . The y-coordinate of Wally is `10*(sqrt 3 + 2)/5 - (1/2)*9.8*((sqrt 3 + 2)/5)^2` = 4.73