Question

hello using the kinematic eqautions for the horizontal projectile of a ball being thrown I am given range of flight dx= 1.02time of flight t= 1.20 sheight of launch dy= 1m I need help finding the initial velocity of the realase

Answer 1

Answer:

intial velocity: 6.02 m /s

Explanation:

To find the initial velocity, we make use of the following kinematic equations

$\begin{gathered} y=y_0+(v_0\sin \theta)t-\frac{1}{2}gt^2 \\ x=(v_0\cos \theta)t \end{gathered}$

Now we know that t = 1.20 s, g = 9.8 m /s, y0 = 1.0 m, and at the ground y =0; therefore, the above equations give

$\begin{gathered} 0=1+(v_0\sin \theta)(1.20)-\frac{1}{2}(9.8)(1.20)^2 \\ 1.02=(v_0\cos \theta)(1.20) \end{gathered}$

solving these equations for v0 sin θ and v0 cos θ gives

$\frac{\frac{1}{2}(9.8)(1.20)^2-1}{\mleft(1.20\mright)}=(v_0\sin \theta)$$\Rightarrow v_0\sin \theta=5.05$

and

$\frac{1.02}{\mleft(1.20\mright)}=(v_0\cos \theta)$$v_0\cos \theta=0.85$

Now dividing the two equations gives

$\frac{v_0\sin \theta}{v_0\cos \theta}=\frac{5.05}{0.85}$$\tan \theta=5.94$

taking the inverse tan gives

$\theta=80.45^o$

Now that we know theta, it is easy to find v0.

$v_0\sin (80.45)=5.94$$v_0=\frac{5.94}{\sin (80.45)}$$\boxed{v_0=6.02}$

which is our answer!