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a cannon ball is launched horizontally from a cannon. the cannon is on the edge of a 50 m tall cliff. how long does the cannon ball take to hit the ground?

User Kensuke
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1 Answer

3 votes

Final answer:

To solve the problem, we can use the kinematic equation for horizontal motion. The cannonball is launched horizontally, so its initial vertical velocity is 0 m/s. Using the equation d = Vit + (1/2)at^2 and substituting the given values, we can solve for time and find that the cannonball takes approximately 3.19 seconds to hit the ground.

Step-by-step explanation:

To solve this problem, we can use the kinematic equation for horizontal motion. The cannonball is launched horizontally, so its initial vertical velocity is 0 m/s. Since it is launched from a 50 m tall cliff, the initial vertical displacement is 50 m. We can use the equation:

d = Vit + (1/2)at^2

Given that the initial vertical velocity is 0 m/s, the acceleration due to gravity is -9.8 m/s^2, and the initial vertical displacement is -50 m (taking downwards as the negative direction), we can substitute these values into the equation and solve for time:

-50 = 0t + (1/2)(-9.8)t^2

Simplifying the equation, we have:

-4.9t^2 = -50

To solve for t, we divide both sides by -4.9:

t^2 = 10.2

t = sqrt(10.2) ≈ 3.19 s

Therefore, the cannonball takes approximately 3.19 seconds to hit the ground.

User Jiayang
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