Consider a cannonball fired from a cannon that rises, reaches a maximum height,

Question

Consider a cannonball fired from a cannon that rises, reaches a maximum height, and falls back down hitting the ground some horizontal distance from the cannon. If you were to fire a second cannonball, leaving the initial vertical speed the same, but increasing the initial horizontal speed, how would the path of the second cannonball differ from the first? The second cannonball would reach a greater maximum height 0 a lower maximum height. the same maximum height. The second cannonball would travel a greater horizontal distance from the cannon. a smaller horizontal distance from the cannon. the same horizontal distance from the cannon.

Explanation / Answer

let Vo be the initial speed and makes an angle theta with horizontal

horizontal component of initial velocity is Vox = Vo*cos(theta)

Vertical component of initial veliocity is Voy = Vo*sin(theta)

maximum height is Hmax = Vo^2*sin^2(theta)/2g

since maximum depends only on vertical component of velocity ,then there is no change in maximum height

so the answer is the same maximum height

horizontal distance travelled is x = Vo*cos(theta)*T

since x depends on Vo*cos(theta)

so the answer is a greater horizontal distance from the cannon

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-> if increaasing the initial vertical speed

then second cannon ball would reach a greater maximum height

the second cannon ball would travel the same horizontal distance from the cannon

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maximum height and horizontal distances does not depend on the mass of the object

hence the answers are

the second cannonball would reach the same maximum height

the second cannon ball would travel

the same horizontal distance from the cannon

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