Question

the maximum horizontal range of a shot from a sling is 250 m find the velocity of projection of the shot what is the height reached by the shot when it has horizontal displacement of 100 root 2 m


Please solve this on the paper please help​

Answer 1

Question:

The maximum horizontal range of a shot from is 250 m. Find the velocity of projection. What is the height reached by the object when it has a horizontal displacement of 100√2 m?

Calculation:

Let initial velocity of projection be "u":

General expression for max range:

$\therefore \: R_{max} = \dfrac{ {u}^{2} }{g}$

$= > \: 250 = \dfrac{ {u}^{2} }{10}$

$= > \: {u}^{2} = 2500$

$\boxed{= > \: u = 50 \: m {s}^{ - 1} }$

Now , for a range of 100√2 , let angle of projection be $\theta$:

$\therefore \: 100 \sqrt{2} = \dfrac{ {u}^{2} \sin(2 \theta) }{g}$

$= > \: 100 \sqrt{2} = \dfrac{ {(50)}^{2} \sin(2 \theta) }{10}$

$= > \: 100 \sqrt{2} = \dfrac{ 2500 \sin(2 \theta) }{10}$

$= > \: 100 \sqrt{2} = 250 \sin(2 \theta)$

$= > \: 10\sqrt{2} = 25 \sin(2 \theta)$

$= > \: \sin(2 \theta) = \dfrac{2 \sqrt{2} }{5}$

$= > \: 2 \theta = {34.4}^{ \circ}$

$= > \: \theta = {17.2}^{ \circ}$

Now , max height :

$\therefore \: h = \dfrac{ {u}^{2} { \sin}^{2}( \theta) }{2g}$

$= > \: h = \dfrac{2500 \: { \sin}^{2}( {17.2}^{ \circ} ) }{20}$

$= > \: h = \dfrac{2500 \times 0.087 }{20}$

$\boxed{ = > \: h = 10.875 \: m}$

HOPE IT HELPS.