0

so i got that , a object was fired from ground with angule $ \theta $. when the object get the height equal to half of his max height , the velocity of object is $\frac{3}{4}$ of initial velocity.Find the initial $\theta$ . i Tryed this way

$$h_{1} = \frac{1}{2}h_{max}$$ $$V_{1} = \frac{3}{4}V_{0}$$ $$V_{1} = V_{0}.cos\theta_{1} \Rightarrow \frac{3}{4}V_{0}= V_{0}.cos\theta_{1} \Rightarrow \frac{3}{4}=cos\theta_{1}$$ $$cos\theta_{1} \approx 48,49^{º}$$

but dont know how to move foward.

ps.sry for the bad english :/

Rogers
  • 149

2 Answers2

1

Let $v_0$ be the initial velocity. Then, $$y=v_0\sin\theta t-\dfrac{1}{2}gt^2\\ y^\prime=v_0\sin\theta-gt=0\\ \implies t_{\mathrm{max}}=\dfrac{v_0\sin\theta}{g}\\ y_{\mathrm{max}}=v_0\sin\theta\dfrac{v_0\sin\theta}{g}-\dfrac{1}{2}g\left(\dfrac{v_0\sin\theta}{g}\right)^2=\dfrac{v_0^2\sin^2\theta}{2g}\\ \implies \dfrac{y_{\mathrm{max}}}{2}=\dfrac{v_0^2\sin^2\theta}{4g}\\ \dfrac{v_0^2\sin^2\theta}{4g}=v_0\sin\theta t_{\mathrm{max}/2}-\dfrac{1}{2}gt_{\mathrm{max}/2}^2$$ Solve for $t_{\mathrm{max}/2}$. Then use $$v=\dfrac{3v_0}{4}=v_0-gt_{\mathrm{max}/2}\\ \implies t_{\mathrm{max}/2}=\dfrac{v_0}{4g}$$ Solve for $\theta$.

1

Recall motion equations for a projectile: \begin{align} x&=(v_0\cos \theta)t, & y&=(v_0\sin \theta)t-\frac{1}{2}gt^2\\ v_x&=v_0\cos\theta,&v_y&=v_0\sin\theta-gt \end{align} Maximun height is reached when $y'=0$, it is $t=\frac{v_0\sin\theta}{g}$ and $h_{\text{max}}=\frac{v_0^2\sin^2\theta}{2g}$. Let $t_1$ the time when $y=\frac{1}{2}h_{\text{max}}$, so \begin{align} (v_0\sin\theta)t_1-\frac{1}{2}gt_1^2=\frac{v_0^2\sin^2\theta}{4g} \end{align} Solving last equation we find two roots, namely $\frac{(2\pm \sqrt{2})v_0\sin\theta}{2g}$, what means there are two instants when $y=\frac{1}{2}h_{\text{max}}$, $t_1=\frac{(2- \sqrt{2})v_0\sin\theta}{2g}$, when projectile is going up, and $t_2=\frac{(2+ \sqrt{2})v_0\sin\theta}{2g}$, when projectile is going down. Then \begin{align} v&=\frac{3}{4}v_0\\ v_0^2\cos^2\theta+v_0^2\sin^2\theta-2(v_0\sin\theta)gt_1+g^2t_1^2&=\frac{9}{16}v_0^2\\ \end{align} Reducing the last equation we get $7-8\sin^2\theta=0$ and then $\sin\theta=\frac{\sqrt{14}}{4}$.

Ángel Mario Gallegos
  • 19,882