A catapult on a cliff launches a large round rock towards a ship on the ocean below. The rock leaves the catapult from a height H = 35.0 m above sea level, directed at an angle theta = 46.9° above the horizontal, and with a speed v = 25.8 m/s. Assuming that air friction can be neglected, calculate the horizontal distance D traveled by the projectile.
I have attempted this 8 times and all of them were wrong.
I have attempted this 8 times and all of them were wrong.
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to find the horizontal distance you need the horizontal velocity and time of flight; you know the horizontal velocity is 25.8 cos 46.9; this stays constant through the flight since there are no horizontal forces acting on the object
so we need to find the time
we find the time by writing the equation of motion in the y direction:
y(t)=35+25.8sin46.9 - 1/2 gt^2
we want to know the time when y=0 (rock hits the surface); set y=0 and get
0=35+25.9sin46.9 t - 4.9t^2
solve this quadratic by the usual means; I get t=5.23s
therefore, the horizontal distance is
dist = speed x time = 25.8cos46.9*5.23s = 92.2m
so we need to find the time
we find the time by writing the equation of motion in the y direction:
y(t)=35+25.8sin46.9 - 1/2 gt^2
we want to know the time when y=0 (rock hits the surface); set y=0 and get
0=35+25.9sin46.9 t - 4.9t^2
solve this quadratic by the usual means; I get t=5.23s
therefore, the horizontal distance is
dist = speed x time = 25.8cos46.9*5.23s = 92.2m