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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.

Felix Marin
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Ella
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    Hint: It will be going up for exactly 1.5 seconds then it will start coming down. Acceleration due to gravity on earth is $9.81 \text{m} \cdot \text{s}^{-2}$ so you can calculate the initial upward velocity. Once you know both the initial vertical and horizontal velocities you can calculate the angle. – Warren Hill Jul 27 '15 at 10:25
  • Hint: Taking g as known, the vertical component of velocity causes zero displacement in 3s. – true blue anil Jul 27 '15 at 10:27
  • Doesn't the acceleration due to gravity (9.8) change due to the initial velocity upwards? – Ella Jul 27 '15 at 10:30
  • Hey, this is very basic. The initial velocity does not change g. If you have been taught to use calculus for finding velocity, use the hint I gave. If you can use any SUVAT equation, find the vertical component of velocity directly using the appropriate equation. – true blue anil Jul 27 '15 at 10:44
  • I have been taught that when breaking the initial velocity into its components, the vertical component velocity effects the acceleration downwards. So I change the acceleration by constructing a right angle triangle with the 9.8 on the hypotenuse and using the complementary of the angle usually given to find the magnitude of the side which makes the angle with the side with magnitude 9.8. This apparently is the new acceleration. It worked before with another problem. Now I am just confused. – Ella Jul 27 '15 at 10:57

2 Answers2

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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.

true blue anil
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  • So... using the right angle triangle I get tan(x)=14.7/25 which gives 30.5 degrees. Is that right? That's what you meant right? (this is for finding the initial angle of projection.) – Ella Jul 27 '15 at 11:38
  • Yes, that's right to one decimal place. – true blue anil Jul 27 '15 at 11:43
  • Thanks for all the help, true blue anil. – Ella Jul 27 '15 at 11:47
  • You're welcome. I understand that giving thanks is to be avoided on this forum; instead, depending on your eligibility, you can upvote and/or tick to accept answer. – true blue anil Jul 27 '15 at 11:54
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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, $\color{blue}{\text{horizontal range}}$ $$\color{blue}{(u\cos \theta)(t)}=\color{blue}{25\times 3=75\ m}$$ $\color{blue}{\text{initial vertical velocity component}}$ $$\color{blue}{u\sin\theta}=\color{blue}{14.7\ m/sec}$$ $\color{blue}{\text{initial angle of projection}\ (\theta)}$ $$\color{blue}{\theta=\tan^{-1}\left(\frac{14.7}{25}\right)\approx 30.45^\circ}$$

Harish Chandra Rajpoot
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