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.