To what height above sea level does the rock rise? This is a tough one, so read

Question

To what height above sea level does the rock rise?

This is a tough one, so read the hint! 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 33.0m 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 135m. Assuming that air friction can be neglected, calculate the value of the angle . (Answer in "deg".) Calculate the speed at which the rock is launched. To what height above sea level does the rock rise?

Explanation / Answer

in the horizontal motion we are taken the velocity asVx as

Vx=displacement/time

Vx = D/t = 135 / 6 = 22.5m/s

in the vertical motion it is due to acceleration due to gravity as velocity Vy here we take in -ve H because we choose the launch point as origin of y

-H = Vyt -(1/2)gt^2

hence Vy = (1/2)gt -H/t = 0.5*9.8*6 - 33/6 =23.9 m/s

tan = Vy/Vx = 22.5/23.9 = 0.9414

= 43.27deg

Vo= squareroot (Vx^2+Vy^2)

   = squareroot(22.5^2+23.9^2)

hence spped of rock is =32.82m/s

but v=u+at

when it reaches max height then initil velocity becomes 0 hence

0 = Vy - gt

t = Vy/g

height from the launch point is

then t value substitute in s=ut+at^2/2 we get

h = Vyt - (1/2)gt^2 = Vy(Vy/g) -(1/2)g(Vy/g)^2 = Vy^2/(2g)

= 22.5^2/(2*9.8) = 25.82meters

hence the max height from the sea level rock rise is

= 33+25.82

= 58.82 meters

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