A cannonball shot at an angle of 60 Degrees had a range of 196m. What was its initial speed?

To find the initial speed of the cannonball, we can use the range formula for projectile motion. The range (R) of a projectile is given by:

R = (V^2 * sin(2θ)) / g

where:
V is the initial speed of the projectile,
θ is the launch angle, and
g is the acceleration due to gravity (approximately 9.8 m/s^2).

In this case, the cannonball has a range of 196m and an angle of 60 degrees. Plugging these values into the range formula, we can solve for the initial speed (V):

196 = (V^2 * sin(2 * 60)) / 9.8

To simplify the equation further, we'll calculate sin(120) since sin(2 * 60) is sin(120):

196 = (V^2 * sin(120)) / 9.8

Using sin(120) = √3 / 2, we can rearrange the equation to solve for V:

V^2 = (196 * 9.8 * 2) / √3
V^2 = 3841.14

Taking the square root of both sides gives us the initial speed (V):

V = √3841.14

Therefore, the initial speed of the cannonball is approximately 62.01 m/s.

L=vₒ²•sin2α/g,

solve for 'v'