Question

At time t=0, a small ball is projected from point A with a velocity of 60 m/s at ${{60}^{\circ }}$ angle with horizontal. Neglect atmospheric resistance and determine the two times ${{t}_{1}}$​ and ${{t}_{2}}$​ where the velocity of the ball makes an angle of ${{45}^{\circ }}$ with the horizontal x− axis.

Answer 1

Hint: There will be two instances when the ball makes an angle of ${{45}^{\circ }}$ with the horizontal x− axis: one in the upward direction and the other in the downward direction. The students should also remember that the velocity in x direction remains the same in both cases.

Complete step by step answer:
In the question we are given that a small ball is projected from point A with a velocity of 60 m/s at ${{60}^{\circ }}$ angle with horizontal. This is done at time t=0.
We have to find the two instances of time where the velocity of the ball makes ${{45}^{\circ }}$ with the horizontal x− axis.

The horizontal component is same in both the points, that is
${{v}_{x}} = u\cos {{60}^{\circ }}=60\times \dfrac{1}{2}=30m/s$
We can compute the vertical component of velocity using the first equation of motion:
${{v}_{y}}=u\sin {{60}^{\circ }}-gt=30\sqrt{3}-10t$
It is said that the velocity makes ${{45}^{\circ }}$ with the horizontal x− axis.
Thus the slope is
$\begin{align}
  & \tan {{45}^{\circ }}=\dfrac{{{v}_{y}}}{{{v}_{x}}} \\
 & \Rightarrow \dfrac{{{v}_{y}}}{{{v}_{x}}}=\pm 1 \\
\end{align}$
This means that
 $\begin{align}
  & {{v}_{y}}=\pm {{v}_{x}}=\pm 30m/s \\
 & \\
\end{align}$
When ${{v}_{y}}=30m/s$,
$\begin{align}
  & 30\sqrt{3}-10{{t}_{1}}=30 \\
 & {{t}_{1}}=3(\sqrt{3}-1)s \\
\end{align}$

When ${{v}_{y}}=-30m/s$,
$\begin{align}
  & 30\sqrt{3}-10{{t}_{2}}=-30 \\
 & {{t}_{2}}=3(\sqrt{3}+1)s \\
\end{align}$
Hence we have obtained both the values of time when the ball makes an angle of ${{45}^{\circ }}$ with the horizontal x− axis.

Note:
Whenever the students solve the problems in projectile motion, they must note the fact that the angles that are used are made with the vertical. The students should also remember that the velocity in x direction remains the same in both cases. If the vertical angle is given in the question, we must subtract it from ${{90}^{\circ }}$.