Question

A batter hits a baseball so that it leaves the bat at speed ${v_0} = 37{\text{ }}\dfrac{m}{s}$ at an angle ${\alpha _0} = {53.1^ \circ }$. Find the position of the ball and its velocity at $t = 2{\text{ }}\sec $.
A. $x = 44.4\;m;{\text{ }}y = 39.6{\text{ }}m$ and $v = 24.4{\text{ }}\dfrac{m}{s}$
B. $x = 44.4\;m;{\text{ }}y = 39.6{\text{ }}m$ and $v = 34.4{\text{ }}\dfrac{m}{s}$
C. $x = 44.4\;m;{\text{ }}y = 59.6{\text{ }}m$ and $v = 24.4{\text{ }}\dfrac{m}{s}$
D. $x = 64.4\;m;{\text{ }}y = 39.6{\text{ }}m$ and $v = 24.4{\text{ }}\dfrac{m}{s}$

Answer 1

Hint: From Second Motion’s equation we have to find out the position of the ball. As we have to find out in $x{\text{ }}$ and $y$ coordinates, we must use the whole equation once in $x$-axis and once in $y$-axis. Later for the velocity of the ball we have found the resultant magnitude of velocity from both $x$-axis and $y$-axis from Motion’s Third equation.

Formula used:
$s = ut + \dfrac{1}{2}a{t^2}$
where $s = $ distance covered, $t = $time, $a = $acceleration, $u = $initial velocity.
$v = u + at$
where $v = $final velocity, $u = $initial velocity, $a = $acceleration and $t = $time.

Complete step by step answer:
From Second equation of motion we get,
$s = ut + \dfrac{1}{2}a{t^2} - - - \left( 1 \right)$
At $x$-axis, $a = 0$, $u = {v_0}\cos {53.1^ \circ }$ and $t = 2{\text{ }}s$
We know that $\cos {53.1^ \circ } = \dfrac{3}{5}$
$s = {v_0}\cos {53.1^ \circ } \times 2 - - - \left( 2 \right)$
Substitution the values in equation $\left( 2 \right)$ we get,
$s = 37 \times \dfrac{3}{5} \times 2 \\
\Rightarrow s = 44.4 \\ $
So, the position of the ball is $x = 44.4{\text{ }}m$.
Now, for the $y$-axis,
$a = g = 9.8{\text{ }}\dfrac{m}{{{s^2}}}$ , $u = {v_o}\sin {53.1^ \circ }$ and $t = 2{\text{ }}s$
Substituting the values in equation $\left( 1 \right)$ we get,
$s = {v_o}\sin {53.1^ \circ } \times 2 - \dfrac{1}{2} \times 9.8 \times {2^2}$
We know that $\sin {53.1^ \circ } = \dfrac{4}{5}$
$s = 37 \times \dfrac{4}{5} \times 2 - \dfrac{1}{2} \times 9.8 \times {2^2} \\
\Rightarrow s = 39.6 \\ $
So, the position of the ball is $y = 39.6{\text{ }}m$.
Now, we have to find the velocity from both $x$-axis and $y$-axis thus find the resultant velocity.
From First equation of motion,
$v = u + at - - - - \left( 3 \right)$
Along $x$-axis let the velocity be $v'$ and $a = 0$ and $u = {v_0}\cos {53.1^ \circ }$.
Substituting the values in equation $\left( 3 \right)$ we get,
$v' = \dfrac{3}{5}{v_0}$
Along $y$-axis let the velocity be $v''$ and $a = g = 9.8{\text{ }}\dfrac{m}{{{s^2}}}$ and $u = {v_0}\sin {53.1^ \circ }$ and $t = 2{\text{ }}s$.
Substituting the values in equation $\left( 3 \right)$ we get,
$v'' = {v_0}\sin {53.1^ \circ } - 9.8 \times 2 \\
\Rightarrow v'' = \dfrac{4}{5}{v_0} - 19.6 \\ $
Thus, the resultant velocity $v = \sqrt {{{\left( {v'} \right)}^2} + {{\left( {v''} \right)}^2}} $
Substituting the values we get,
$v = \sqrt {{{\left( {22.2} \right)}^2} + {{\left( {10} \right)}^2}} \\
\therefore v = 24.4 $
The velocity of the ball after $t = 2{\text{ }}\sec $ is $24.4{\text{ }}\dfrac{m}{s}$.

So, the correct option is A.

Note: We must resolve the equation into two parts of $x$-axis and $y$-axis. The velocity of the ball should also be the resultant velocity of both components. Most use a single component and solve which is wrong. Acceleration due to gravity must be $g = 9.8{\text{ }}\dfrac{m}{{{s^2}}}$ not round off value or else the answer will not be correct.