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 35.0 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 148.0 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

here,

let the initial speed be v0 and the angle be theta

height , H = 35 m

time taken , t = 6 s

hprizontal distance , R = 148 m

vo*cos(theta) * t = 148

v0 * cos(theta) = 24.667 ....(1)

and

- H = v0*sin(theta)*t - 0.5 * g*t^2

- 35 = v0*sin(theta)* 6 - 0.5 * 9.8* 6^2

v0 * sin(theta) = 23.57 ...(2)

from first and seccond equation

theta = 43.7 degree

v0 = 34.12 m/s

height above the cliff be h'

0 - ( v0 * sin(theta))^2 = - 2 * g * h

h = 28.35 m

the height above the sea level ( 28.35 + 35)

the height above the sea level is 63.35 m

the speed at which the rock is launched is 34.12 m/s   

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