A04. Calculate the maximum height of the object above the ground below the cliff

Question

A04. Calculate the maximum height of the object above the ground below the cliff.

A05. Find the velocity vector in polar form at the instant when the object hits the ground

A03 through A05. A projectile is launched from a cliff and strikes the ground below as shown in the diagram. You are given the initial speed and direction of the projectile, and the height of the cliff, but not the (horizontal) range vo = 60 m/s 60° A03. Calculate the time of flight of the projectile in seconds 200 m Ranges ?

Explanation / Answer

a) Hmax = h + voy^2/(2*g)

= h + (vo*sin(thet))^2/(2*g)

= 200 + (60*sin(60))^2/(2*9.8)

= 337.8 m

b) x-component of velocity of projectile when it hits the ground, vx = vo*cos(60) = 60*cos(60) = 30 m/s

let vy is the component of velocity.

Apply^2, Vy^2 - Voy^2 = 2*g*h

vy = -sqrt(voy^2 + 2*g*h)

= -sqrt((60*sin(60))^2 + 2*9.8*200)

= -81.4 m/s

so, v = sqrt(vx^2 + vy^2)

= sqrt(30^2 + 81.4^2)

= 85.75 m/s

theta = tan^-1(vy/vx)

= tan^-1(-81.4/30)

= 290.2 degrees with +x axis in counetr closkwise direction.

so, v = ( 85.75 m/s, 290.2 degrees)

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