Object launched at an angle from a height

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Unit: Work, Energy, and Power

Prerequisites

Energy Conservation for Launch Problems

Later Topics

None

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Part 1

A small block of mass $m$ starts from rest at the top of a frictionless track at height $H = 4.24\,\mathrm{m}$ above level ground. The track dips down and then rises into a straight ramp. The end of the ramp is at height $h = 0.81\,\mathrm{m}$ above the ground and makes an angle of $\theta = 28^\circ$ above the horizontal. The block leaves the ramp and moves as a projectile. Neglect air resistance. Take $g = 9.8\,\mathrm{m/s^2}$ and $U_g = 0$ at ground level.

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Using conservation of mechanical energy from the top of the track (height $H$ ) to the end of the ramp (height $h$ ), derive a formula for the speed $v_0$ of the block as it leaves the ramp in terms of $g$ , $H$ , and $h$ . Then calculate its numerical value.

Correct!

Solution:

Solution

We apply conservation of mechanical energy between the initial position at the top of the track and the point where the block leaves the ramp.

At the top of the track (height $H$ ), the block starts from rest, so: $E_{\text{initial}} = mgH$

At the end of the ramp (height $h$ ), the block has speed $v_0$ : $E_{\text{final}} = \frac{1}{2}mv_0^2 + mgh$

By conservation of energy: $mgH = \frac{1}{2}mv_0^2 + mgh$

Dividing through by $m$ and rearranging: $gH = \frac{1}{2}v_0^2 + gh$

$\implies g(H - h) = \frac{1}{2}v_0^2$

$\implies v_0 = \sqrt{2g(H - h)}$

Substituting the given values $g = 9.8\,\mathrm{m/s^2}$ , $H = 4.24\,\mathrm{m}$ , and $h = 0.81\,\mathrm{m}$ :

$v_0 = \sqrt{2 \cdot 9.8 \cdot (4.24 - 0.81)} = 8.2\,\mathrm{m/s}$

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