Question

A projectile is launched with an initial speed of $30\dfrac{m}{s}$ at an angle of $60^\circ$ above horizontal. What are the (a) magnitude and (b) angle of its velocity 2s after launch and (c) is the angle above or below the horizontal. What are the (d) magnitude and (e) angle of its velocity 5s after launch and (f) is the angle above or below the horizontal.

Answer 1

When an object is thrown with some initial velocity at some angle with the horizontal then this type of motion is known as projectile motion. The maximum range of the projectile is when angle from the horizontal is $45^\circ$.

Formula used:
The velocity of the projectile at any instant is given by,
$\vec v = {v_o}\cos {\theta _o}\hat i + \left( {{v_o}\sin {\theta _o} - gt} \right)\hat j$
Where initial velocity is ${v_o}$ and the angle from the horizontal is${\theta _o}$.

Complete step by step solution:
It is given in the problem that a projectile with initial velocity $30\dfrac{m}{s}$ at an angle of $60^\circ$ and we need to find (a) magnitude and (b) angle of its velocity 2s after launch and (c) is the angle above or below the horizontal. What are the (d) magnitude and (e) angle of its velocity 5s after launch and (f) is the angle above or below the horizontal.
The velocity of the projectile at any instant is given by,
$\Rightarrow \vec v = {v_o}\cos {\theta _o}\hat i + \left( {{v_o}\sin {\theta _o} - gt} \right)\hat j$
Where initial velocity is ${v_o}$ and the angle from the horizontal is${\theta _o}$.
The value ${v_o} = 30\dfrac{m}{s}$ and ${\theta _o} = 60^\circ$ at t=2s.
$\Rightarrow \vec v = {v_o}\cos {\theta _o}\hat i + \left( {{v_o}\sin {\theta _o} - gt} \right)\hat j$
$\Rightarrow \vec v = 30\cos 60^\circ \hat i + \left( {30 \times \sin 60^\circ - 9. 8 \times 2} \right)\hat j$
$\Rightarrow \vec v = \left( {30 \times \dfrac{1}{2}} \right)\hat i + \left( {30 \times \dfrac{{\sqrt 3 }}{2} - 9. 8 \times 2} \right)\hat j$
$\Rightarrow \vec v = 15\hat i + \left( {\dfrac{{15\sqrt 3 }}{2} - 19. 6} \right)\hat j$
$\Rightarrow \vec v = 15\hat i + 6. 4\hat j$
The magnitude of the velocity is equal to,
$\Rightarrow \left| {\vec v} \right| = \sqrt {{{15}^2} + {{\left( {6. 4} \right)}^2}}$
$\Rightarrow \left| {\vec v} \right| = \sqrt {225 + 40. 96}$
$\Rightarrow \left| {\vec v} \right| = \sqrt {265. 96}$
$\Rightarrow \left| {\vec v} \right| = 16\dfrac{m}{s}$
(b) the direction of $\vec v$.
$\Rightarrow \theta = {\tan ^{ - 1}}\left( {\dfrac{{6. 4}}{{15}}} \right)$
$\Rightarrow \theta = {\tan ^{ - 1}}\left( {0. 43} \right)$
$\Rightarrow \theta = 23^\circ$.
(c) The angle is positive and therefore the angle is above horizontal.
(d) The velocity of projectile at t=5s is equal to,
$\vec v = {v_o}\cos {\theta _o}\hat i + \left( {{v_o}\sin {\theta _o} - gt} \right)\hat j$
$\vec v = 30 \times \cos 60^\circ \hat i + \left( {30 \times \sin 60^\circ - 9. 8 \times 5} \right)\hat j$
$\vec v = 30 \times \dfrac{1}{2}\hat i + \left( {30 \times \dfrac{{\sqrt 3 }}{2} - 49} \right)\hat j$
$\vec v = 15\hat i + \left( {15\sqrt 3 - 49} \right)\hat j$
$\vec v = 15\hat i - 23\hat j$
(d)The direction of velocity is equal to,
$\Rightarrow \theta = {\tan ^{ - 1}}\left( { - \dfrac{{23}}{{15}}} \right)$
$\Rightarrow \theta = - 57^\circ$
(e) The angle is in clockwise direction and therefore it is below the horizontal.

Note: The students are advised to understand and remember the formula for velocity of projectile at any instant. The angle is positive if the angle is taken anti-clockwise then it is negative. The projectile frame is taken as the positive x and y axis.