a toy cannon fires a projectile from ground level with a speed of 15.0 m/s, at an angle of 30.0 degrees above the horizontal.
a) calculate the horizontal range of projectile.
R=Vi^2/g * sin 2theta
= (150/9.8)*(sin(2*30.0)
= 19.88m
I know this part is right but i'm not sure what to do next?
b) calculate the maximum height for the trajectory.
c) calculate the components of the projectile's impact velocity.
R=vₒ²•sin2α/g,
h= vₒ²•sin²α/2g,
t= 2vₒ•sinα/g
v(x) =v₀(x)= v₀•cos α
v(y) =-v₀(y) = - v₀•sin α
To calculate the maximum height reached by the projectile, you can use the equation:
h = (V^2 * sin^2(theta)) / (2 * g)
where:
- h is the maximum height
- V is the initial velocity (15.0 m/s)
- theta is the launch angle in radians (30.0 degrees converted to radians is 0.524 radians)
- g is the acceleration due to gravity (9.8 m/s^2)
Plugging these values into the equation, we have:
h = (15.0^2 * sin^2(0.524)) / (2 * 9.8)
= (225 * 0.27316) / 19.6
= 3.4527 meters
So, the maximum height reached by the projectile is approximately 3.45 meters.
To calculate the components of the projectile's impact velocity, you can use the equations:
- The horizontal component (Vx): Vx = V * cos(theta)
- The vertical component (Vy): Vy = V * sin(theta)
Plugging in the values, we have:
Vx = 15.0 * cos(0.524)
= 15.0 * 0.866
= 12.99 m/s (rounded to two decimal places)
Vy = 15.0 * sin(0.524)
= 15.0 * 0.5
= 7.50 m/s (rounded to two decimal places)
So, the horizontal component of the projectile's impact velocity is approximately 12.99 m/s, and the vertical component is approximately 7.50 m/s.