Question Find the horizontal range of a projectile launched at 15 degrees to the horizontal with speed of 40m/s

To Find : The horizontal range of a projectile launched at 15 degrees to the horizontal with speed of 40 m/s. Solution : The horizontal range of a projectile is given by : [tex]R = \dfrac{u^2 sin 2\theta}{g}[/tex] ( Here, g is acceleration due to gravity = 10 m/s² ) Putting all value in above equation : [tex]R = \dfrac{40^2 \times sin (2 \times 15)}{10} \ m\\\\R = \dfrac{1600 \times 1}{2\times 10} \ m\\\\R = 80 \ m[/tex] Therefore, the horizontal range of projectile is 80 m. Log in to Reply

To Find :The horizontal range of a projectile launched at 15 degrees to the horizontal with speed of 40 m/s.

Solution :The horizontal range of a projectile is given by :

[tex]R = \dfrac{u^2 sin 2\theta}{g}[/tex] ( Here, g is acceleration due to gravity = 10 m/s² )

Putting all value in above equation :

[tex]R = \dfrac{40^2 \times sin (2 \times 15)}{10} \ m\\\\R = \dfrac{1600 \times 1}{2\times 10} \ m\\\\R = 80 \ m[/tex]

Therefore, the horizontal range of projectile is 80 m.