Physics- projectile motion. Given values: time of flight and horizontal component velocity. Ball launched at an angle returning to the same height.

Question

A person throws a baseball with a horizontal component velocity of $25$ m/s: It takes $3$ seconds to come back to its original height.

• Calculate -friction is ignored-
  • its horizontal range,
  • its initial vertical component velocity and
  • its initial angle of projection.

I have found the horizontal range: $3 \times 25 = 75\ \mbox{m}$, but I can't find other two questions.

I find that I seem to not have enough information to solve the next question.

Answer 1

g, you will recall, is known as the gravitational constant. [It is the vertical component of velocity that keeps on changing due to the effect of g during the flight of the projectile ]

Using the notation $U_x$ for horizontal component of velocity and $U_y$ and $V_y$ for initial and final vertical component of velocity, and taking the upward direction as positive,

$V_y = U_y - 9.8\times1.5$

At its maximum height, $V_y = 0$, thus $U_y = 9.8\times1.5 =14.7$ m/s

You already know that $U_x = 25$ m/s

Now construct your right angle triangle and proceed.

Answer 2

Edit: The acceleration due to Earth's gravity, $g=9.8\ m/sec^2$

Let, $u$ be the velocity of projectile & $\theta$ be the angle of projection. Now, we have $$\text{horizontal velocity component}=\color{blue}{u\cos \theta=25}\tag 1$$ $$\text{vertical velocity component}=u\sin \theta$$ Since, it takes $3\ sec$ to come back to the initial height hence it should take $t=1.5\ sec$ to reach to the maximum height from point of projection where the final vertical velocity component becomes zero. Now using First Equation of Motion as follows $$V=U+at$$ For motion against gravity, we take negative value of gravitational acceleration $g$ as follows $$0=u\sin\theta-gt$$ $$\implies u\sin\theta-(9.8)(1.5)=0 \implies \color{blue}{u\sin\theta=14.7}\tag 2$$ Now, diving (2) by (1), we get $$\frac{u\sin\theta}{u\cos\theta}=\frac{14.7}{25}\implies \color{blue}{\tan\theta=\frac{14.7}{25}} $$ Hence, $\color{blue}{\text{horizontal range}}$ $$\color{blue}{(u\cos \theta)(t)}=\color{blue}{25\times 3=75\ m}$$ $\color{blue}{\text{initial vertical velocity component}}$ $$\color{blue}{u\sin\theta}=\color{blue}{14.7\ m/sec}$$ $\color{blue}{\text{initial angle of projection}\ (\theta)}$ $$\color{blue}{\theta=\tan^{-1}\left(\frac{14.7}{25}\right)\approx 30.45^\circ}$$