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 152.0 m. Assuming that air friction can be neglected, calculate the value of the angle ?.

Calculate the speed at which the rock is launched.

To what height above sea level does the rock rise?

Explanation / Answer

1. Work out the horizontal velocity (this doesn't change during the flight). So it's pretty easy: it's just 152m/6s. =25.33 m/s

2.
Basically:
h=vt - (gt^2)/2
-32 = v.6s - (g.(6s)^2)/2
And solve for v. =24.06

3. Now you've got the initial horizontal and vertical velocities, it should be easy to work out the angle and initial velocity using trig and pythagoras.

tan(theta)=24.06/5.33=43.52 deg

Total speed=sqrt(u^2+v^2)=34.93 m/s

4. H=32+u^2/2g=32+24.06^2/2g=61.53 m

where u is the horizontal velocity

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