Answer to Question #160619 in Molecular Physics | Thermodynamics for Nike

Question

A projectile is fired from ground level, with velocity 50 m/s at 30 degrees to the horizontal. Find its horizontal range, the greatest vertical height to which it rises, and the time to reach the greatest height. Hence determine the least speed with which it could be projected in order to achieve the same horizontal range? ( resistance of air to the motion is negligible)

Answer 1

$v_0 = 50\,\mathrm{m/s}, \alpha = 30^\circ.$

Let x-axis be directed horizontally and y-axis be directed vertically.

$v_x = v_0\cos\alpha, \;\; v_y = v_0\sin\alpha- gt.$

The horizontal range can be calculated as

$L = \dfrac{2v_0^2\sin\alpha\cos\alpha}{g} = \dfrac{v_0^2\sin2\alpha}{g} = 216\,\mathrm{m}.$

The time of flight is $t = 2\dfrac{v_0\sin\alpha}{g} = 5\,\mathrm{s}.$

The greatest height is reached after 2.5 seconds of flight.

$h = v_0\sin\alpha (t/2) - \dfrac{g(t/2)^2}{2} = 31.3\,\mathrm{m}.$

From the formula of the horizontal range we may see that the greatest range for the constant speed corresponds to the maximum $\sin2\alpha = 1,$ so $\alpha = 45^\circ.$ Therefore, if we want to minimize the initial velocity we should maximize $\sin2\alpha.$

$v_1 = \sqrt{\dfrac{Lg}{1}} = 46.5\,\mathrm{m/s}.$