a caNon standing on a flat field fires a cannonball with a muzzle velocity v, at

Question

a caNon standing on a flat field fires a cannonball with a muzzle velocity v, at an angle theta above horizontal. The cannonball thus initially has velocity components vx=vcostheta and vy=vysintheta. A) show that the cannons range(horizontal distance to where the cannonball falls) is given by the equation R=(2v^2/g)sintheta costheta. B) interpret your equation in the cases theta=0 and theta=90 degrees. C) show the maximum range is for theta=45 degrees. a caNon standing on a flat field fires a cannonball with a muzzle velocity v, at an angle theta above horizontal. The cannonball thus initially has velocity components vx=vcostheta and vy=vysintheta. A) show that the cannons range(horizontal distance to where the cannonball falls) is given by the equation R=(2v^2/g)sintheta costheta. B) interpret your equation in the cases theta=0 and theta=90 degrees. C) show the maximum range is for theta=45 degrees. a caNon standing on a flat field fires a cannonball with a muzzle velocity v, at an angle theta above horizontal. The cannonball thus initially has velocity components vx=vcostheta and vy=vysintheta. A) show that the cannons range(horizontal distance to where the cannonball falls) is given by the equation R=(2v^2/g)sintheta costheta. B) interpret your equation in the cases theta=0 and theta=90 degrees. C) show the maximum range is for theta=45 degrees.

Explanation / Answer

A)

Here , for initial velcoity Vy in vertical direction

time of flight , t = 2 * Vy/g

t = 2 * v * sin(theta)/g

Now , range , R = t * Vx

R = 2 * v * sin(theta)/g * (v * cos(theta))

R = 2 * v^2 & sin(theta) * cos(theta)/g

B)

for theta = 0

as sin(theta) = 0

R = 0

for theta = 90 degree

cos(theta) = 0

R = 0

C)

as R = 2 * V^2 * sin(theta) * cos(theta)/g

R = V^2 * sin(2*theta)/g

for maximum range , sin(theta ) = 1

2 * theta = 90

theta = 45 dehree

hence , Range is maximum for theta = 45 degree

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