A man stands on the roof of a building of height y 0and throws a rock with a vel

Question

A man stands on the roof of a building of height y0and throws a rock with a velocity of magnitude v0at an angle of above the horizontal. You can ignore air resistance.

Part A:

Calculate the maximum height above the roof reached by the rock.

Take free fall acceleration to be g

Part B:

Calculate the magnitude of the velocity of the rock just before it strikes the ground.

Take free fall acceleration to be g

Part C:

Calculate the horizontal distance from the base of the building to the point where the rock strikes the ground.

Take free fall acceleration to be g

Explanation / Answer

As given in the question,

Initial velocity = u = v0, angle = , Height of building = h = y0, acceleration = a = -g

=>   u(x) = v0 cos and u(y) = v0 sin

(A) For the maximum height above the roof reached by the rock,

v^2 = u^2 + 2*a*s , where v = 0 at the topmost position

=> 0 = (v0 sin)^2 + 2*(-g)*s =>   s = (v0 sin)^2 / 2*g

(B) For the magnitude of the velocity of the rock just before it strikes the ground,

Total height from the ground, H = h + s = y0 + (v0 sin)^2 / 2*g

Again using,         v^2 = u^2 + 2*a*s , where u = 0 at the topmost position

=> v(y)^2 = 2*g*H = 2*g*{y0 + (v0 sin)^2 / 2*g}

As there is no acceleration in x-direction => v(x) = u(x) = v0 cos

                 => v(x)^2 = (v0 cos)^2

So, the magnitude of final velocity when it is about to reach the ground,

v = sqrt[v(x)^2 + v(y)^2] = sqrt[(v0 cos)^2 + 2*g*{y0 + (v0 sin)^2 / 2*g}]

(C) For the horizontal distance from the base of the building to the point where the rock strikes the ground,

We first calculate the total time elapsed using:

  v = u + a*t

                => v(y) = u(y) + (-g)*t

                => t = [u(y) - v(y)] / g = [v0 sin - sqrt(2*g*{y0 + (v0 sin)^2 / 2*g})] / g

So the horizontal distance,

   d = u(x)*t = v0 cos * [u(y) - v(y)] / g = [v0 sin - sqrt(2*g*{y0 + (v0 sin)^2 / 2*g})] / g

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