A cannon mounted on a pirate ship fires a cannonball at 125 m/s horizontally, at a height of 17.5 m above the ocean surface. Ignore air resistance. (a) How much time elapses until it splashes into the water? (b) How far from the ship does it land?
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http://www.jiskha.com/display.cgi?id=1292451447
Same problem. Different numbers
To find the answers to these questions, we can use basic kinematic equations and the equations of motion.
(a) How much time elapses until it splashes into the water?
To find the time it takes for the cannonball to splash into the water, we can use the equation of motion for vertical motion:
h = ut + (1/2)gt^2
where:
h is the height (17.5 m),
u is the initial vertical velocity (0 m/s),
g is the acceleration due to gravity (-9.8 m/s^2),
t is the time we want to find.
Since the initial vertical velocity is 0 m/s (the cannonball is only fired horizontally), the equation simplifies to:
h = (1/2)gt^2
Plugging in the values, we have:
17.5 = (1/2)(-9.8)t^2
Simplifying further:
-19.6t^2 = 17.5
Dividing both sides by -19.6:
t^2 = -17.5 / -19.6
t^2 = 0.8929
Taking the square root of both sides:
t ≈ √0.8929
t ≈ 0.945 seconds
Therefore, it takes approximately 0.945 seconds for the cannonball to splash into the water.
(b) How far from the ship does it land?
To find the horizontal distance the cannonball travels, we can use the equation of motion for horizontal motion:
s = ut
where:
s is the horizontal distance we want to find,
u is the initial horizontal velocity (125 m/s),
t is the time taken (0.945 seconds).
Plugging in the known values:
s = 125 * 0.945
s ≈ 118.125 meters
Therefore, the cannonball lands approximately 118.125 meters from the ship.