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 34.0 m above sea level, directed at an angle theta above the horizontal with an unknown speed v_s. The projectile remains in flight for 6.00 seconds and travels a horizontal distance D of 168.0 m. Assuming that air friction can be neglected, calculate the value of the angle theta. You know how long the rock is in the air. This yields its horizontal velocity from the horizontal distance it travels. You get the vertical component of the velocity by using the equation of motion with constant acceleration. From the two velocity components you can figure out the angle. Calculate the speed at which the rock is launched. To what height above sea level does the rock rise?

Explanation / Answer

along horizontal

speed = v0x = x/t = 168/6 = 28

along vertical

y = v0y*t + (1/2)*ay*t^2

-34 = v0y*6 - (1/2)*9.8*6^2

v0y = 23.7 m/s

tantheta = v0y/v0x

tantheta = 23.7/28

theta = 40.2 degrees <<<-----------answer

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speed v0 = sqrt(v0x^2+v0y^2)

speed v0 = sqrt(28^2+23.7^2) = 36.8 m/s <<<========answer

++++++++++++++++++++++++++++++++

at the maximm height vy = 0

vy^2 - v0y^2 = 2*ay*dy

0 - 23.7^2 = -2*9.8*(y-34)

y = 62.66 m <<<------------answer

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