Question

A boy can throw a stone up to a maximum height of 10m. The maximum horizontal distance that the boy can throw the same stone up to will be:
$\begin{aligned} & A.\quad 20\sqrt{2}m \\\ & B.\quad 10m \\\ & C.\quad 10\sqrt{2}m \\\ & D.\quad 20m \\\ \end{aligned}$

Answer 1

We need to find the values of different parameters for a projectile motion of a body initially fired at an angle of $\theta$with the horizontal. We will then find the maximum height of the projectile: ${{h}_{\max }}=\dfrac{{{u}^{2}}{{\sin }^{2}}\theta }{2g}$. The range (R) of the projectile motion is, $R=\dfrac{{{u}^{2}}\sin 2\theta }{g}$.

Step by step solution:
Let’s start by mentioning the information that is given to us. We are given that the maximum height reached by the stone is 10m. Hence, ${{h}_{\max }}=10m$. Therefore, in this case, for the stone to reach maximum height, the stone must have been thrown vertically upward to achieve the maximum height as given.
Hence in this case, the initial velocity of the stone (u), will not have any horizontal component. That is, if the initial velocity of the projectile is (u) makes an angle of ${{\theta }^{0}}$with the horizontal. We will break the initial velocity (u) into the horizontal component given by: ${{u}_{x}}=u\cos \theta$. Similarly, the vertical component of the initial velocity (u) is given by: ${{u}_{y}}=u\sin \theta$.
And in this case, the angle $\theta ={{90}^{0}}$, Hence, ${{u}_{y}}=u\sin 90=u$and ${{u}_{x}}=u\cos 0=0$.
To find the maximum height of the projectile, we will consider the newton’s laws of motion in two dimension: ${{h}_{\max }}=\dfrac{v_{y}^{2}-u_{y}^{2}}{2g}$. At the maximum height of the projectile motion, the final velocity of the body along the vertical will be zero and the acceleration due to gravity will be (-g), since it is against the direction of the motion. Therefore: ${{h}_{\max }}=\dfrac{v_{y}^{2}-u_{y}^{2}}{2g}\Rightarrow 10=\dfrac{(0)_{{}}^{2}-(u)_{{}}^{2}}{2(-g)}\Rightarrow 10=\dfrac{{{u}^{2}}}{2g}\Rightarrow u=\sqrt{20g}$.
We know that the horizontal range, that is the horizontal distance covered by a projectile during projectile motion is given by the formula: $R=\dfrac{{{u}^{2}}\sin 2\theta }{g}$, for maximum horizontal range: $R={{R}_{\max }}\Rightarrow \sin 2\theta =1\Rightarrow \theta ={{45}^{0}}$. Hence, the maximum horizontal range formula becomes: $R=\dfrac{{{u}^{2}}}{g}$. Substituting in the value of the initial velocity (u) from above, we get the value of range (R) as: ${{R}_{\max }}=\dfrac{{{u}^{2}}}{g}\Rightarrow {{R}_{\max }}=\dfrac{{{\left( \sqrt{20g} \right)}^{2}}}{g}=20m$

Note: We must remember that at maximum height, the vertical velocity of the projectile will become zero, and external acceleration of gravity acts on the particle. For the horizontal motion, there isn’t any external acceleration acting on it. Hence, the initial and final horizontal velocity of the projectile, will always be constant in a projectile motion.