You are hired to run the circus cannon which has an adjustable muzzle speed v 0

Question

You are hired to run the circus cannon which has an adjustable muzzle speed v0 and angle of inclination . You need to specify these parameters in order to shoot a clown onto a platform 10 m away horizontally and 10 m high (relative to the muzzle end) while reaching a maximum height of 11 m. (a) Calculate the vertical velocity component required to reach a maximum height of 11 m. Hint: to do this, combine the vertical velocity and position equations to eliminate time. (b) Calculate the time required to reach maximum height assuming t = 0 is when the clown leaves the muzzle. (c) Calculate the time required to fall from 11 m to 10 m. (d) Using the answers to parts (b) and (c), calculate the horizontal velocity component required for the clown to reach the platform. (e) From the answers to parts (a) and (d), calculate the muzzle speed and the angle of inclination of the cannon. (f) Write the initial velocity of the clown as a vector.

Explanation / Answer

(a)   If we know that the equations mentioned in the hint are:

y = v0y*t - (0.5)*g*t^2     and    vy = v0y - g*t; and we also know that at the maximum height vy = 0, we can get:

v0y = (2*g*ymax)^(0.5) = (2*9.8*11)^(0.5) [m/s] = 14.7 [m/s]

(b) The answer is to calculate the maximum time, which is as follow:

tmax = v0y/g = 14.7/9.8 [s] = 1.5 [s]

(c) As they are telling us that the clown (the object) is falling, we must assume that we have passed the maximum height, so we should calculate the time from that point to y = 10 m and then add the maximum time. To put it simple, we are at the second half of the movement.

y = y0 + v0y*t - (0.5)*g*t^2;   in this case y = 10 m , y0 = 11 m , v0y = 0   so :   t1 = (2*(y0 - y)/g)^(0.5)

t1 = (2*(11 - 10)/9.8)^(0.5) [s] = 0.45 [s] ; now, we add the time of the first half of the movement

tv = t1 + tmax = 0.45 + 1.5 [s] = 1.95 [s]

(d) To find the answer we only have to remember that the velocity in the x axis is constant so:

x = v0x*tv; which means; v0x = x/tv = 10/1.95 [m/s] = 5.13 [m/s]

(e)   Up to this point we have v0y and v0x. We know that they, together with v0, form a right triangle and v0 is the hypotenuse. So, using the Pythagorean theorem:

v0 = (v0x^2 + v0y^2)^(0.5) = (5.13^2 + 14.7^2)^(0.5) [m/s] = 15.6 [m/s]

Finally, if we remember how the tangent is (as we have the legs of the triangle), we can find the angle:

angle = tg^-1 (v0y/v0x) = tg^-1 (14.7/5.13) = 70.8 °

(f) The initial velocity as a vector is:   v = 5.13 i + 14.7 j

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