A projectile is thrown in the upward direction making an angle of \(60^{\circ}\) with the horizontal with a…
A projectile is thrown in the upward direction making an angle of \(60^{\circ}\) with the horizontal with a velocity of \(140 \mathrm{~ms}^{-1}\). Then the time after which its velocity makes an angle \(45^{\circ}\) with the horizontal is (Acceleration due to gravity, \(g=10 \mathrm{~ms}^{-2}\) )
\(0.5124 \mathrm{~s}\)
\(51.24 \mathrm{~s}\)
\(5.124 \mathrm{~s}\)
\(512.4 \mathrm{~s}\)
Solution
Given, angle of projection, \(\theta=60^{\circ}\) and velocity,
\(u=140 \mathrm{~ms}^{-1}\)
This velocity have divided into two components, horizontal component, \(u_x=u \cos 60^{\circ}=140 \cos 60^{\circ}\) and vertical component, \(u_y=u \sin 60^{\circ}=140 \sin 60^{\circ}\) Let after time \(t\), the inclination of particle with horizontal be \(45^{\circ}\) and at time \(t\) velocity along \(x=v_x\) and along \(y=v_y\).
Now,
\(\begin{aligned}
\tan 45^{\circ} & =\frac{v_y}{v_x} \\
& =v_x=v_y
\end{aligned}\)
Since, the horizontal component of velocity remains constant, i.e.,
\(\begin{aligned}
v_x & =u_x=140 \cos 60^{\circ} \\
\text{and } v_y & =u_y-g t \\
140 \cos 60^{\circ} & =140 \sin 60^{\circ}-10 t \quad\left[\because v_y=v_x\right] \\
140 \times \frac{1}{2} & =140 \times \frac{\sqrt{3}}{2}-10 t \\
70 & =70 \sqrt{3}-10 t \\
10 t & =70 \sqrt{3}-70 \\
t & =\frac{70(\sqrt{3}-1)}{10} \\
& =7 \times(0.7320)=5.124 \mathrm{~s}
\end{aligned}\)