A catapult on a cliff launches a large round rock towards a ship on the ocean be

Question

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 of 32.0 m above sea level, directed at an angle ?, above the horizontal with an unknown speed v0.The projectile remains in flight for 6.00 seconds and travels a horizontal distance D of 140.0 m. Assuming that air friction can be neglected, 1) calculate the value of the angle ?, 2) Calculate the speed at which the rock is launched, 3)To what height above sea level does the rock rise? please explain what formulas you derive your work from

Explanation / Answer

a)

From the given we can write that
y(6)=0=32+v0*sin(?)*6-.5*g*36
and
x(6)=140=v0*cos(?)*6

rearrange the equations

18*g-32=v0*sin(?)*6
140=v0*cos(?)*6

divide
tan(?)=(18*g-32)/140
45.89 degrees above the horizontal

b)

speed of rock = 140/(6cos45.89) = 33.52 m/s

c)

v = v0 sin?-g*t=0

t = v0sin?/g =

vertical distance travelled

y = 32m+v0sin?*v0sin?/g-1/2*g*(v0sin?/g)^2

= 32+v0^2sin^2?/2g

by substituting values we get

y = 61.553 m

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