Question

The Horizontal eange of a projectile fired at an angle of 15° is 50 m , if it is fired with same speed at an angle of 45° it's range will be .

Answer 1

Heya........!!!!

___________________

Let the angle of projection be ( alpha )
= Initial velocity = u
= Acceleration due to gravity = g

$in \: question \\ \alpha \: = \: 15 \\ r \: = 50 \: m \\ \\ = > \: range \: = \frac{u {}^{2} \sin(2 \alpha ) }{g} \\ puttin \: the \: values \\ \\ = > \: \: 50 = \frac{u {}^{2} \sin(30) }{9.8 } \\ = > u {}^{2} = 980 \\ = > u = \sqrt{980} \\ = > u = 31.304 \frac{m}{s} \\ \\ now \: \alpha \: = 45 \\ = > \: r = \frac{u {}^{2} \sin(2 \times 45) }{g} \\ = > r \: \: = \frac{u {}^{2} }{g} \: \: \: ( \sin(90) \: \: = \: 1) \\ \\ r \: = \: \frac{(14 \sqrt{5} ) {}^{2} }{9.8} \\ = > \: r \: = \: 100 \: m$


=============================

Hope It Helps You. ☺

Answer 2

Answer:

R₂ = 99.85 m

Explanation:

It is given that the horizontal range of a projectile fired at an angle of 15° is 50 m. The horizontal range of a projectile is defined as the horizontal displacement. Mathematically, it can be written as :

$R=\dfrac{v^2sin\ 2\theta}{g}$

v is the initial speed of the projectile

$50=\dfrac{v^2sin\ 2(30)}{10}$

v = 31.6 m/s...........(1)

Second condition, if it is fired with same speed at an angle of 45°. Its range will be :

$R_2=\dfrac{(31.6)^2sin\ 2(45)}{10}$

$R_2=99.85\ m$

Hence, if the projectile is fired with a speed of 31.6 m/s, its range will be 99.85 m.