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 31.0 m above sea level, directed at an angle above the horizontal with an unknown speed v0.

1-The projectile remains in flight for 6.00 seconds and travels a horizontal distance D of 142.0 m. Assuming that air friction can be neglected, 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?

Explanation / Answer

a)

From the given graph, we can write that
y(6) = 0 = 31 + v0*sin()*6 - 0.5*g*36
and
x(6) = 142 = v0*cos()*6

rearrange the equations

18*g - 31 = v0*sin()*6
142 = v0*cos()*6

divide
tan() = (18*g - 31)/142

tan() = (18*9.8 - 31)/142

tan() = 1.024

= 45.67 degrees above the horizontal

b)

speed of rock, v = 142 / (6cos 45.67)

                      v = 33.87 m/s

c)

v = v0 sin - g*t = 0

t = v0sin/g

vertical distance travelled

y = 31m + v0sin*v0sin/g - 1/2*g*(v0sin/g)^2

   = 31+ v0^2*sin^2 / 2g

by substituting values we get

y = 31+ 33.87^2*sin^2 45.67 / 2*9.8

y = 60.95 m

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