You and your cannon arc located at the top of a cliff that has a height of 550 m

Question

You and your cannon arc located at the top of a cliff that has a height of 550 m. Your cannon fires shells with an initial velocity of 380 m/s. The target is on a plateau that has an elevation of 270 m and the horizontal distance from your location to the target is 525 m. (a) What is the minimum angle at which you could point your cannon in order to hit the target? (b) How long did it take to get there? (c) What is the speed of the shell when it arrives? (d) What is the maximum angle at which you could point your cannon and again hit the target? (e) How long does it take to get there along this path? (f) Now, what is its speed when it arrives?

Explanation / Answer

Horizontal distance travlled is x = Vo*cos(theta)*t = 525 m

vertical distance travelled is y = (Vo*sin(theta)*t) - (0.5*g*t^2)

Vo*cos(theta)*t = 525

t = 525/(Vo*cos(theta))

therefore

y = [Vo*sin(theta)*(525/(Vo*cos(theta)))] + (0.5*g*(525/(Vo*cos(theta)))^2) = 270

(525*tan(theta)) + (0.5*9.8*(525/(380*cos(theta)))^2) = 270

theta = 27.06 degrees

b) t = 525/(380*cos(27.06)) = 1.56 sec

C) Vx= Vox = Vo*cos(theta) = 380*cos(27.06) = 338.4 m/sec

Vy = Voy + (g*t) = (Vo*sin(theta)) - (g*t)

Vy = (380*sin(27.06)) + (9.8*1.56) = 188.15 m/sec

v = sqrt(338.4^2+188.15^2) = 387 m/sec

range is R = Vo^2*cos^2(theta)/(2*g) = 525

380^2*cos^2(theta)/(2*9.8) = 525

cos^2(theta) = 525*19.6/380^2 = 0.0712

cos(theta) = sqrt(0.0712)= 0.266 or -0.266

theta = 74.5 degrees

D) theta_max= 74.5 degrees

e)) t = R/(Vo*cos(theta)) = 525 /(380*cos(74.5)) = 5.16 sec

f) Vx = Vox = 380*cos(74.5) = 101.55 m/sec

Vy = Voy+ (g*t) = Vo*sin(theta) + (g*t)

Vy = (380*sin(74.5)) + (9.8*5.16) = 416.74 m/sec

V = sqrt(101.55^2+416.74^2) = 428.93 m/sec

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