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202211241532 Solution to 2019-DSE-PHY-IA-6

A small ball after projection moves under the effect of gravity only. Its velocity at a certain instant is shown below. What is the speed of the ball 1\,\mathrm{s} before? Neglect air resistance. (g=9.81\,\mathrm{m\,s^{-2}})

This figure is not one original but a modified.

Roughwork.

Assume the angle of projection be \theta (t_0) =\theta_0 with speed v(t_0)=v_0 at launch time t=t_0 such that the resolved x-, y– components

\begin{aligned} v_x(v_0,\theta_0 ) & = v_0\cos\theta_0 \\ v_y(v_0,\theta_0 ,t) & = v_0\sin\theta_0 -gt \\ \end{aligned}

are combined to give the resultant speed at certain instant t:

\begin{aligned} v(v_0,\theta_0,t) & =\sqrt{\big(v_x(v_0,\theta_0)\big)^2+\big(v_y(v_0,\theta_0,t)\big)^2} \\ & = \sqrt{v_0^2-2v_0(\sin\theta_0)gt+g^2t^2} \\ \end{aligned}

making with the level angle \theta (t)

\begin{aligned} \theta (v_0,\theta_0,t) & = \arctan\bigg( \frac{v_y(v_0,\theta_0,t)}{v_x(v_0,\theta_0)}\bigg) \\ & = \arctan\bigg( \frac{v_0\sin\theta_0-gt}{v_0\cos\theta_0} \bigg) \\ \end{aligned}

Now that for some instant t=T we are given

\begin{aligned} \theta (v_0,\theta_0,T) & =180^\circ -90^\circ -27^\circ =63^\circ \\ v(v_0,\theta_0,T) & =11\,\mathrm{m\,s^{-1}} \\ \end{aligned}

recall

\displaystyle{y(x) = \bigg(-\frac{g}{2v_0^2\cos^2\theta_0}\bigg) x^2+(\tan\theta_0)x}

Equating

\begin{aligned} \tan\theta \big|_{\theta =63^\circ} & = \frac{\mathrm{d}y}{\mathrm{d}x}\bigg|_{(x_0,y_0)} \\ & = \bigg(-\frac{g}{v_0^2\cos^2\theta_0}\bigg)x_0 +\tan\theta_0 \\ \end{aligned}

Differentiating on \theta (v_0,\theta_0,t) wrt time t:

\begin{aligned} \frac{\mathrm{d}\theta}{\mathrm{d}t} & = \frac{\mathrm{d}}{\mathrm{d}t}\bigg[\arctan \bigg(\frac{v_0\sin\theta_0-gt}{v_0\cos\theta_0}\bigg)\bigg] \\ & = \frac{1}{1+\Big(\frac{v_0\sin\theta_0-gt}{v_0\cos\theta_0}\Big)^2} \\ \theta'(t) & = \frac{v_0^2\cos^2\theta_0}{v_0^2-2v_0(\sin\theta_0)gt+g^2t^2} \\ \end{aligned}

and also on v(v_0,\theta_0,t):

\begin{aligned} \frac{\mathrm{d}v}{\mathrm{d}t} & = \frac{\mathrm{d}}{\mathrm{d}t}\bigg( \sqrt{v_0^2-2v_0\sin\theta_0gt+g^2t^2}\bigg) \\ v'(t) & = \frac{v_0(\sin\theta_0)g+g^2t}{\sqrt{v_0^2-2v_0(\sin\theta_0)gt+g^2t^2}} \\ & = \frac{v_0(\sin\theta_0)g+g^2t}{v_0\cos\theta_0} \cdot \sqrt{\theta'(t)} \\ \end{aligned}

(to be continued)