Question

A cannonball is launched from level ground at an angle of elevation of 55 degrees and observe to impact level ground downrange at a distance of 432 m. The same cannon is then used to launch a cannonball from level ground at an angle of elevation of 38 degrees into the side of a cliff and the cannonball is observed to strike the cliff 28.5 m above level ground.

A) How far is the base of the cliff from the cannon (measured along level ground)?

B) What maximum height does the cannonball reack above level ground?

C) What is the impact velocity of the cannonball with the cliff? (state your answer in i-j notation.)

Answer 1

Let vo is the initial velocity of the canon ball.

R = vo^2*sin(2*theta)/g

vo = sqrt(R*g/sin(2*theta))

= sqrt(432*9.8/sin(2*55))

= 67.1 m/s

A) when projected at 38 degrees.

vox = vo*cos(38) = 67.1*cos(38) = 52.88 m/s

voy = vo*sin(38) = 67.1*sin(38) = 41.31 m/s

let t is the time taken to hit th cliff.

h = voy*t - 0.5*g*t^2

28.5 = 41.31*t - 4.9*t^2

4.9*t^2 - 41.31*t + 28.5 = 0

on solving the above eqiation we get

t = 7.67 s

so, x = vox*t

= 52.88*7.67

= 405.6 m <<<<<<----------Answrr

b) Hmax = voy^2/(2*g)

= 41.31^2/(2*9.8)

= 87 m <<<<<<----------Answrr

c) vy = voy - g*t

= 41.31 - 9.8*7.67

= -33.8 m/s

so, v = vx i + vy j

= (52.88 i - 33.8 j )m/s <<<<<<----------Answrr