Find the range of a projectile launched at an angle of 45 with an intial velocity of 25 m/s .
Angle 45degree and velocity 25 m/what is range
Find answer
R = v^2/g
Studying
Range=(initial velocity)^2 times sin(2teta)/g. =(25m/s)^2 x sin(90)/9.81 =63.7meter
To find the range of a projectile launched at an angle of 45 degrees with an initial velocity of 25 m/s, you can use the range formula for projectile motion.
The range formula is given as:
Range (R) = (V^2 * sin(2θ)) / g
Where:
- V is the initial velocity of the projectile
- θ is the launch angle of the projectile
- g is the acceleration due to gravity (approximately 9.8 m/s^2)
In this case, V = 25 m/s and θ = 45 degrees.
First, convert the launch angle from degrees to radians:
θ_in_radians = θ * (π/180)
θ_in_radians = 45 * (π/180)
θ_in_radians = π/4
Now, substitute the values into the range formula:
R = (25^2 * sin(2 * (π/4))) / 9.8
Simplify the equation:
R = (625 * sin(π/2)) / 9.8
R = (625 * 1) / 9.8
R = 625 / 9.8
R ≈ 63.77 meters
Therefore, the range of the projectile is approximately 63.77 meters.